Copyright © 2012 Carlson, O’Bryan & Joyner Worksheet #1: Introduction to Quadratic Functions...

Worksheet #1: Introduction to Quadratic Functions

Worksheet #2: Properties of Quadratic Functions

Worksheet #3: Modeling Quadratic Relationships

Worksheet #4: Zeros (Roots) of Quadratic Functions

MODULE 4Quadratic Functions

Part 1: Worksheets #1-5

Worksheet #5: Factoring Quadratic Functions

1. Gina wanted to explore the idea of speeding up with her students. She walked across the front of the room, traveling 1 foot over the 1st second, 3 feet over the 2nd second, 5 feet over the 3rd second, 7 feet over the 4th second and 9 feet over the 5th second.

INTRODUCTION TO QUADRATIC FUNCTIONS

a. Complete the table to show how far Gina traveled over each of the following intervals.

IntervalDistance Gina travels

during the interval (feet)

from 0 to 1 second

from 1 to 2 seconds

from 2 to 3 seconds

from 3 to 4 seconds

INTRODUCTION TO QUADRATIC FUNCTIONS

a. Complete the table to show how far Gina traveled over each of the following intervals.

IntervalDistance Gina travels

during the interval (feet)

from 0 to 1 second 1

from 1 to 2 seconds 3

b. Do you notice a pattern to how Gina’s distance traveled over each interval changes? If so, describe the pattern in your own words.

Over each 1-second interval she traveled 2 feet further than she did during the previous 1-second interval.

c. Explain how you know that Gina’s speed is increasing (called acceleration).

She travels a greater distance over equal intervals of time as the time since she began walking increases, so her speed must be increasing. d. If Gina continues the same pattern of acceleration, how

far will she travel over the next two 1-second intervals?She will travel 11 feet from 5 to 6 seconds since she began walking, and she will travel 13 feet from 6 to 7 seconds since she began walking.

e. Construct a table of the total distance Gina traveled d (measured in feet) in terms of the amount of time t (measured in seconds) since she started walking. (Assume that Gina stops walking after 7seconds.)

Number of seconds since Gina began

walking (t)

Total distance (in feet) Gina

has traveled (d)

Distance Gina travels during

the interval (∆d)

walking (t)

has traveled (d)

the interval (∆d)

5 2511

6 3613

walking (t)

has traveled (d)

the interval (∆d)

Change in ∆d

5 2511

6 3613

walking (t)

has traveled (d)

the interval (∆d)

Change in ∆d

1 1 23

2 4 25

3 9 27

4 16 29

5 25 211

6 36 213

f. How are the number of seconds since Gina began walking (t) and total distance Gina has traveled (d) related for integer values of t? Give the formula for d in terms of t.

The total distance Gina has traveled since she began walking is the square of the number of seconds she has been walking. d = t2 g. Construct a graph of Gina’s total distance traveled (in

feet) in terms of the number of seconds since she started walking for integer values of t. Pay attention to the domain and range based on the context.

h. On your graph, show the change in the number of seconds elapsed from t = 0 to t = 1, then show the corresponding change in the total distance Gina has traveled. Repeat this for the intervals from t = 1 to t = 2, t = 2 to t = 3, t = 3 to t = 4, and t = 4 to t = 5. Discuss how the graph supports the idea that Gina is accelerating.

Gina travels a greater distance over each 1-second interval than she did in the previous 1-second interval, meaning she her speed must be increasing as the time since she began walking increases.

i. Use the formula you developed in part (f) to fill in the tables below showing the total distance Gina had traveled for non-integer values of t.

Number of seconds since Gina began walking (t)

has traveled (d)

0 0 0 0

0.5 0.25

1 1 0.5

1.5 0.75

2 4 1 1

2.5 1.25

3 9 1.5

3.5 1.75

has traveled (d)

0 0 0 0

0.5 0.25 0.25 0.0625

1 1 0.5 0.25

1.5 2.25 0.75 0.5625

2 4 1 1

2.5 6.25 1.25 1.5625

3 9 1.5 2.25

3.5 12.25 1.75 3.0625

j. Calculate Gina’s change in distance over each ½-second interval (for the table on the left) and each ¼-second interval (for the table on the right), then determine how ∆d is changing. What pattern do you notice in how Gina’s distance increases for equal increases in time since she started walking?

has traveled (d)∆d

Change in ∆d

0.5 0.25

1.5 2.25

2.5 6.25

3.5 12.25

Change in ∆d

0 00.25

0.5 0.250.75

1 11.25

1.5 2.251.75

2 42.25

2.5 6.252.75

3 93.25

3.5 12.25

Change in ∆d

0 00.25

0.5 0.25 0.500.75

1 1 0.501.25

1.5 2.25 0.501.75

2 4 0.502.25

2.5 6.25 0.502.75

3 9 0.503.25

3.5 12.25

Change in ∆d

0.25 0.0625

0.5 0.25

0.75 0.5625

1.25 1.5625

1.5 2.25

1.75 3.0625

Change in ∆d

0 00.0625

0.25 0.06250.1875

0.5 0.250.3125

0.75 0.56250.4375

1 10.5625

1.25 1.56250.6875

1.5 2.250.8125

1.75 3.0625

Change in ∆d

0 00.0625

0.25 0.0625 0.1250.1875

0.5 0.25 0.1250.3125

0.75 0.5625 0.1250.4375

1 1 0.1250.5625

1.25 1.5625 0.1250.6875

1.5 2.25 0.1250.8125

1.75 3.0625

The function in Exercise #1 is an example of a quadratic function. Quadratic functions have the characteristic that the change in the output values follows a linear pattern for equal changes in the input. The example in Exercise #1 (d = t2) is the simplest case of a quadratic function. In this case, when ∆t = 1, then ∆d increases by 2 additional feet each second.

has traveled (d)

Change in the total distance

Gina has traveled (∆d)

Change in ∆d

has traveled (d)

Change in ∆d

5 2511

6 3613

has traveled (d)

Change in ∆d

1 1 23

2 4 25

3 9 27

4 16 29

5 25 211

6 36 213

x y ∆y Change in ∆y

14 101

Another example of a quadratic function is shown below. In this case, when ∆x = 2 we see that ∆y increases by 4 more for each 2-unit change in x.

4 1110

6 2114

8 3518

10 5322

12 7526

14 101

2 5 46

4 11 410

6 21 414

8 35 418

10 53 422

12 75 426

14 101

0 –2

Compare to a linear function, such as y = 3x – 2.

0 –26

10 286

12 346

0 –26

2 4 06

4 10 06

6 16 06

8 22 06

10 28 06

12 34 06

2. y = 2x2 + 8

In Exercises #2-5, use the given formula of a quadratic function to complete the table, then find the pattern in the changes in y for each change in x.

3 2614

4 40 418

5 58 422

6 80 426

7 106 430

3. y = –3x2

1 –3–24

3 –27 –24–48

5 –75 –24–72

7 –147 –24–96

9 –243 –24–120

11 –363

4. y = x2 + 4x

–2 –41

–1 –3 23

0 0 25

1 5 27

2 12 29

5. y = 2(x + 4)2

–11 98–66

–8 32 36–30

–5 2 366

–2 8 3642

1 50 3678

6. (left table)

6. Examine each of the following tables and explain why they do not represent quadratic functions of y with respect to x.

–4 2

–1 8

6. (left table)

–4 26

–1 8 410

0 18 414

3 32 418

4 50 422

The relationship might appear to be quadratic at first. However, on closer inspection we see that ∆x is not consistent, so the fact that the change in ∆y is constant in this table doesn’t mean anything.

6. (right table)

6. Examine each of the following tables and explain why they do not represent quadratic functions of y with respect to x.

6. (right table)

1 1 67

2 8 1219

3 27 1837

4 64 2461

We see that ∆x is consistent, but the change in ∆y isn’t constant, so the relationship can’t be quadratic.

No, he is incorrect. His conclusions rely on reasoning that is only true for linear functions. From x = 3 to x = 4 the change in y will be greater than from x = 2 to x = 3. Our classmate is mistaken in thinking that ∆y is the same when x changes from 2 to 3 and when x changes from 3 to 4.

7. A classmate was examining the following table that shows some ordered pairs for a quadratic function. He says that if x = 3, then y = 13. Do you agree? Defend your position.

8. The graph of a quadratic function is called a parabola. Examine the patterns of change of the output while considering equal increments of change of the input for each of the following graphs. Which, if any, could represent the graph of a quadratic function?

a. –11

a. x y ∆y Change in ∆y

–2 –615

–1 9 –141

0 10 –2–1

1 9 –14–15

2 –6

This can’t represent a quadratic function of y with respect to x because, for equal changes in x, the change in ∆y isn’t constant.

–2 7–6

0 1 4–2

2 –1 42

4 1 46

This appears to represent a quadratic function of y with respect to x because, for equal changes in x, the change in ∆y is constant.

0 –3–1

1 –4 –2–3

2 –7 –2–5

3 –12

This could represent a quadratic function of y with respect to x because, for equal changes in x, the change in ∆y is constant. (However, more ordered pairs would allow us to be more confident.)

1 2 12

2 4 24

3 8 48

This appears to represent a quadratic function of y with respect to x because, for equal changes in x, the change in ∆y is constant.

9. The following table shows ordered pairs for a quadratic function. In the table, we see that as x increases, y increases. How do we know that, eventually, y must decrease?

Quadratic functions are different from linear functions in that (when we examine their entire possible domain) they have an interval where the output quantity increases and an interval where the output quantity decreases. This has to do with the pattern of changes that we’ve explored.

Since ∆y changes by –2 whenever x increases by 1, eventually ∆y must become negative. When ∆y is negative, the function is decreasing as x increases.

Let’s continue the table to see this occur.

5 –2–1

–2–2

10. The following table shows ordered pairs for a quadratic function. In the table, we see that as x increases, y decreases. How do we know that, eventually, y must increase?

Since ∆y changes by 8 whenever x increases by 1, eventually ∆y must become positive. When ∆y is positive, the function is increasing as x increases.

Let’s continue the table to see this occur.

–12–20

–74 3

–86–90

In Exercises #11-12, fill in the table so that the relationship between y and x forms a quadratic function.

Answers will vary. This is only one of many possible responses.

In Exercises #11-12, fill in the table so that the relationship between y and x forms a quadratic function.

Answers will vary. This is only one of many possible responses.

13. Graph the points (x, y) from the table in Exercise #11. Then show the change in x and change in y between each point.

Again, answers will vary. We draw the graph that matches our sample answer to Exercise #13.

14. Explain (in your own words) what you’ve learned about quadratic functions in this worksheet.

1. Which of the following tables could represent ordered pairs from a quadratic function?

PROPERTIES OF QUADRATIC FUNCTIONS

This table can represent a quadratic function of y with respect to x.

This table does not represent a quadratic function of y with respect to x. (This appears to represent a linear function of y with respect to x.)

This table does not represent a quadratic function of y with respect to x.

f. Pick two of the tables (in (a) through (e) above) that you believe represent values from a quadratic function. Add one more row to each of these tables and predict the y value that corresponds to the new x-value in your table.

–8–44

–144

12 134

2. In Worksheet #1 we explored the growth pattern of f (t) = t2 where d = f (t) for values of t ≥ 0 by examining how the quantities, time elapsed t (in seconds) and Gina’s distance traveled (in feet) since she started moving, varied together.

a. Graph function f and show the changes in distance traveled over equal changes in time on the interval 0 ≤ t ≤ 7.

b. On what interval(s) is f increasing?

f is increasing on the entire interval 0 < t ≤ 7

c. A graph is said to be concave up (or to have positive concavity) if the change in the output variable is increasing for successive equally-sized intervals of the input (in this case t). On what interval(s) is the graph of f concave up? Justify your answer.

f is concave up (or has positive concavity) on the entire interval 0 ≤ t ≤ 7.

We can see this best by revisiting the table we created in Worksheet #1, which shows that the change in ∆d is positive.

walking (t)

has traveled (d)

the interval (∆d)

Change in ∆d

1 1 23

2 4 25

3 9 27

4 16 29

5 25 211

6 36 213

3. Let’s take the function from Exercise #2 out of its original context (i.e., consider f (x) = x2), which removes the restriction that the input values can’t be negative.

a. Complete the third column of the table showing the changes in the output values of the function as x increases by increments of one unit on the interval

–5 ≤ x ≤ 5.

b. Use the fourth column to track the change in ∆y as x increases by increments of 1 unit.

c. On what interval(s) is f increasing? On what interval(s) is f decreasing?

d. On what interval(s) is f concave up (positive concavity)? Which column in the table shows us the concavity of the function most clearly?

f is increasing on the interval 0 < x ≤ 5 and decreasing on the interval –5 ≤ x < 0

The fourth column shows the concavity the most clearly.Since ∆y always changes by 2 as x increases by 1, thefourth column is a positive constant. The function is concave up on the entire interval –5≤ x ≤ 5.

Previously we defined what it means for a function to be concave up. Functions can also be concave down. If x represents values of the input quantity for a function and y represents values of the output quantity, then

• a function is concave up (or has positive concavity) on an interval if, for equal increases in x, ∆y is increasing.

• a function is concave down (or has negative concavity) on an interval if, for equal increases in x, ∆y is decreasing.

4. The following table of values shows a function (g(x) = –2x2 + 5) that is concave down (has negative concavity). Complete the third and fourth columns for the table and explain in your own words why this function is concave down.

This function is concave down because the values of ∆y are decreasing as x increases by equal increments. (The change in ∆y is –1 when ∆x = 0.5.)

5. Sketch the graph of a function (it doesn’t necessarily have to be a quadratic function) over an interval where the function is concave down. Illustrate on your graph how you know the function is concave down on this interval.

For each given quadratic function in Exercises #6-11, do the following. i) Complete the table by filling in the missing outputs and showing ∆y and the changes in ∆y. ii) State the interval(s) over which the function is increasing (if any), then state the interval(s) over which the function is decreasing (if any). iii) State the interval(s) over which the function has positive concavity (if any), then state the interval(s) over which the function has negative concavity (if any). iv) Sketch the function’s graph.

6. y = x2 – 4x + 2

ii. Increasing on 2 < x ≤ 6, decreasing on –1 ≤ x < 2

iii. Positive concavity on –1 ≤ x ≤ 6, no negative concavity

6. y = x2 – 4x + 2

7. y = –2(x + 1)(x + 5)

ii. Increasing on –5 ≤ x < –3, decreasing on –3 < x ≤ 2

iii. Negative concavity on –5 ≤ x ≤ 2, no positive concavity

7. y = –2(x + 1)(x + 5)

8. y = (x – 4)(x + 1)

ii. Increasing on 1.5 < x ≤ 2.5, decreasing on –1 ≤ x < 1.5

iii. Positive concavity on –1 ≤ x ≤ 2.5, no negative concavity

8. y = (x – 4)(x + 1)

9. y = –x2 – 7

ii. Increasing on –6 ≤ x < 0, decreasing on 0 < x ≤ 8

iii. Negative concavity on –6 ≤ x ≤ 8, no positive concavity

9. y = –x2 – 7

10. y = –4(x + 2)2 + 8

ii. Increasing on –3 ≤ x < –2, decreasing on –2 < x ≤ –1.25

iii. Negative concavity on –3 ≤ x ≤ –1.25, no positive concavity

10. y = –4(x + 2)2 + 8

11. y = 3(x – 4)2 + 6

ii. Increasing on 4 < x ≤ 22, decreasing on 1 ≤ x < 4

iii. Positive concavity on –1 ≤ x ≤ 22, no negative concavity

11. y = 3(x – 4)2 + 6

13. The vertex of a parabola is either a maximum or a minimum based on whether the output value of the vertex is the largest or smallest value of y possible for the function. Review Exercises #6-11 and determine if the vertex is a maximum or minimum in each case.

6) minimum 7) maximum 8) minimum 9) maximum 10) maximum 11) minimum

12. Every quadratic function has a vertex. The vertex is the point where the function values change from increasing to decreasing or from decreasing to increasing as the input values increase (as we move from left to right on the graph). What is the vertex for each of the functions in Exercises #6- 11?

6) (2, –2) 7) (–3, 8) 8) (1.5, –6.25) 9) (0, –7) 10) (–2, 8) 11) (4, 6)

15. What is the relationship between the axis of symmetry and the vertex of a parabola?

The axis of symmetry passes through the vertex. If the axis of symmetry is x = h, then the x -coordinate of the vertex is h (and vice versa).

14. The U-shaped graph of a quadratic function is called a parabola. (Caution! Not all U-shaped graphs are parabolas. Review Worksheet 1 to see an example of a U-shaped graph that is not a parabola.) All parabolas can be folded across a vertical line such that the two sides of the parabola match up exactly. The vertical line about which the parabola is symmetric is called the axis of symmetry. What is the axis of symmetry for each of the functions in Exercises #6-11?

6) x = 2 7) x = –3 8) x = 1.5 9) x = 0 10) x = –2 11) x = 4

16. The concavity of a quadratic function is always positive (concave up) or always negative (concave down). How can we tell by looking at the graph what the concavity of a quadratic function is?

Knowing the concavity can tell us whether the vertex is a minimum or a maximum. Positive concavity indicates that the vertex is a minimum. Negative concavity indicates that the vertex is a maximum.

17. If we know the concavity of a quadratic function, what does this tell us about the vertex of the parabola?

If the parabola opens up, we know the concavity must be positive. If the parabola opens down, we know the concavity must be negative.

18. A partial table of values is given below for two quadratic functions. The function represented by the table on the left has a vertex at (–1, 9), and the function represented by the table on the right has a vertex at (4, 2).

a. What is the axis of symmetry for each function?

x = –1 and x = 4

18. A partial table of values is given below for two quadratic functions. The function represented by the table on the left has a vertex at (–1, 9), and the function represented by the table on the right has a vertex at (4, 2).

b. Complete the table of values for each quadratic function.

c. Is each function concave up or concave down?

The function represented by the table on the left is concave down (has negative concavity).

c. Is each function concave up or concave down?

The function represented by the table on the right is concave up (has positive concavity).

d. Is the vertex of each function a maximum or a minimum?

For the function represented by the table on the left, the vertex is a maximum.

For the function represented by the table on the right, the vertex is a minimum.

1. A rancher has 400 meters of fencing. He will fence off a rectangular corral next to the wall of a cliff that will form the north side of the enclosure. So, he will use all 400 meters to make just 3 sides of the rectangle (the west, south, and east sides). He wants to make a corral that will hold as many cattle as possible, so he wants the corral to contain the largest area possible.

MODELING QUADRATIC RELATIONSHIPS

Use the Corral Applet to answer the following questions.

a. As the length of the west side of the corral increases from 0 to 20 meters how does the length of the south side of the corral change?

The length of the south side of the corral changes from 200 meters to 190 meters.

b. As the length of the west side of the corral increases by any amount, how does the length of the south side of the corral change?

The length of the south side changes by –1/2 times as much as the change in the length of the west side of the corral.

c. As the length of the west side of the corral increases from 0 to 20 meters, how does the combined length of the north and south sides of the corral change?

d. As the length of the west side of the corral increases by any amount, how does the combined length of the north and south sides of the corral change?

The combined length of the north and south sides of the corral decrease by 20 meters.

As the length of the west side of the corral increases by any amount, the combined length of the north and south sides of the corral decreases by the same amount.

e. How does the length of the west side of the corral relate to the combined length of the north and south sides?

The length of the west side of the corral plus the combined length of the north and south sides of the corral must be 400 meters. Put another way, the length of the west side of the corral is the difference between the total amount of fencing (400 meters) and the combined length of the north and south sides of the corral.

f. As the west side of the corral increases, how does the total amount of fencing used change?

The total amount of fencing used does not change regardless of the length of the west side of the corral.

g. As the west side of the corral increase from 0 to 10 meters, how does the area of the corral change?

The area contained within the corral increases from 0 square meters to 1,950 square meters (a change of 1,950 square meters).

h. As the west side of the coral increases from 10 to 20 meters, how does the area of the corral change?

i. How long should the rancher build the south side and west side of the corral so that its area is as large as possible?

The area contained within the corral increases from 1,950 square meters to 3,800 square meters (a change of 1,850 square meters).

Exploring this with the applet, we find that the greatest area occurs when the west length is 200 meters and the south length is 100 meters.

2. A manufacturing engineer has been given the job of designing an aluminum container having a square base and rectangular sides to hold screws and nails. It also must be open at the top. The container must be 5 inches tall, but the engineer can make the base whatever size he feels is appropriate.

a. Sketch a diagram of this container. Let x represent the length of the side of the square base in inches.

b. Suppose the outside of the box will be painted. What is the surface area of the box that needs to be painted if the side of the square base is 9 inches? 4 inches? x inches?

9-inch sides for the base4[(9)(5)] + [(9)(9)]

4(45) + 81261 square inches

4-inch sides for the base4[(4)(5)] + [(4)(4)]

4(20) + 1696 square inches

x-inch sides for the base4[(x)(5)] + [(x)(x)]

4(5x) + x2

x2 + 20x square inches

c. Use a calculator to graph the function that shows the surface area (in square inches) if the length of the side of the square base is x inches.

i) What is the vertex, axis of symmetry, and horizontal intercepts for this function?

The vertex is at (–10, –100), the axis of symmetry is x = –10, and the horizontal intercepts are (–20, 0) and (0, 0).

ii) Are these features in the function’s domain? Explain.

The domain of the function is x > 0 [although one might argue that the domain is x ≥ 0] because the length of a side of a box must be positive. Therefore, with the possible exception of (0, 0), none of these features fall within the function’s domain.

d. What is the formula for the volume (in cubic inches) of the container as a function of the length of the side of the square base?

Let V be the volume of the box in cubic inches and V = h(x). h(x) = 5x2

e. What is the volume of the container if the square base has sides that are 6.5 inches long?

h(6.5) = 5(6.5)2 = 211.25 cubic inches

f. What must the dimensions of the base be if the volume of the container is 423.2 cubic inches? Give two methods for finding these dimensions.

The base must be 9.2 inches by 9.2 inches.

One method is to graph h(x) = 5x2 and V = 423.2 and find the intersection point(s) with a calculator. The second method is algebraic.

3. Suppose a baseball outfielder fields a ball and throws it back towards the infield releasing it from his hand 6.5 feet above ground level at an angle of 18o above the horizontal at a speed of 103 feet per second. Neglecting air resistance, the baseball’s height above the ground h (in feet) after t seconds since it was released can be modeled by the function f (t) = –16t2 + 31.829t + 6.5 where h = f (t).

a. Fill in the given table and use the information to draw a graph of the function h.

*Note that t = 2.4 falls outside the domain if this function since it produces heights for the ball below ground level. We include the values here for the sake of being thorough, but entries marked with a * can just as easily be left blank or marked “N/A”.

b. Explain what information you can gather about the height of the baseball as the time since it left the player’s hand increases.

As the number of seconds since the ball was thrown increases, the height of the ball above the ground in feet initially increases, reaches a maximum height, and then decreases towards the ground.

Prior to reaching its maximum height, the height of the ball above ground is increasing, but by smaller and smaller amounts for equal changes in t. After the maximum height, the ball’s height above the ground is decreasing, and it falls further and further for equal changes in t.

Throughout the entire time the ball is in the air, ∆h is always decreasing.

c. The function’s axis of symmetry is approximately t = 0.995. i) What is the baseball’s maximum height above the ground?

f (0.995) = –16(0.995)2 + 31.829(0.995) + 6.5 ≈ 22.33

The ball’s maximum height is about 22.33 feet.ii) Over what time interval(s) is the height of the ball

above ground increasing? Decreasing?

The ball’s height above the ground is increasing between 0 seconds and 0.995 seconds since the ball was thrown. The ball’s height above the ground is decreasing between 0.995 seconds since it was thrown and the time when the ball is caught or it hits the ground.

iii) If the ball is caught by an infielder at a height of 6.5 feet above ground level, how long did it take for the ball to reach the infielder from the time the outfielder threw it?

This question is easiest to answer using the axis of symmetry.

The ball was thrown from a height of 6.5 feet above the ground. If it’s caught at the same height, then it takes the same amount of time to travel from the thrower’s hand to its maximum height as it does to travel from its maximum height to the player’s glove who caught the ball.

Therefore, the ball is caught when t = 0.995(2) = 1.99.

iv) Provide two values of t for which the height will be the same. Explain your method for answering this question.

Answers will vary. The easiest way to answer this is to simply take values of t equidistant from t = 0.995.

For example, at

t = 0.995 – 0.2 = 0.795 and

t = 0.995 + 0.2 = 1.195

we know the ball must have the same height above the ground.

d. Using your graph, solve the equation f (t) = 19 for t (your answer(s) will be approximate). What does the solution represent in the context of this problem?

h = 19

d. Using your graph, solve the equation f (t) = 19 for t (your answer(s) will be approximate). What does the solution represent in the context of this problem?

We estimate that f (t) = 19 when t ≈ 0.55 and t ≈ 1.45. This means that the ball is 19 feet above ground approximately 0.55 and 1.45 seconds since it was thrown.

4. If an object starts at rest and accelerates at a rate of a feet per second each second then the distance the object travels (in feet) after t seconds is given by the quadratic function f (t) = (1/2)at2.

a. Suppose a car accelerates at a rate of 14 feet per second each second over the course of 8 seconds.

i) Explain what this statement means.The car’s speed increases by 14 ft/sec for each second the car accelerates. So after 1 second the car’s speed is 14 ft/sec, after 2 seconds the car’s speed is 28 ft/sec, and so on.

ii) Write the formula used to calculate the car’s distance traveled (in feet) after t seconds spent accelerating.

f (t) = (1/2)(14)t2 f (t) = 7t2

4. If an object starts at rest and accelerates at a rate of a feet per second each second then the distance the object travels (in feet) after t seconds is given by the quadratic function f (t) = (1/2)at2.

a. Suppose a car accelerates at a rate of 14 feet per second each second over the course of 8 seconds.

iii) How far did the car travel during the 8 seconds it spent accelerating?

f (8) = 7(8)2 f (8) = 448

The car traveled 448 feet while accelerating.

b. Suppose another car travels at a constant speed for the same 8 second-period. How fast must this car be traveling if it travels the same distance as the accelerating car?This second car must travel 448 feet in 8 seconds at a

constant speed, so it’s speed must be4488

___ = 56 feet per second

c. We’ve learned that the graph of a quadratic function is symmetric. If we graph the distance traveled for this car over 8 seconds, however, we don’t see a graph with symmetry. Why not?

Within the domain of this function we only have the graph of half of a parabola. The other half (that would by symmetric to the first half) is outside of the function’s domain.

5. A company makes bolts, screws, and washers. The washers are made with holes of different sizes to fit the various size bolts and screws according the following diagram. (For this exercise, recall that the area of a circle A is given by A = πr2 where r is the circle’s

radius.)

a. What is the radius of the outside circle of the washer?

x + 0.8 cm

b. What is the area of the hole in the washer if the radius of the hole is x cm? π(x)2 square centimeters

c. What is the area of the washer’s top surface if the radius of the circular hole is 0.5 cm?

area of the larger circle – area of the holeπ(0.5 + 0.8)2 – π(0.5)2

1.69π – 0.25π144π

The area is 1.44π (or about 4.52) square centimeters

d. Write a formula that calculates the area of the washer’s top surface if the radius of the circular hole is x cm.

Let A be the area of the washer’s top surface (in square cm), and A = f (x). f (x) = π(x + 0.8)2 – π(x)2

1. Consider the function f (x) = (2x – 5)(x + 3). This function is written in factored form because the formula is written as a product of factors, namely the factors (2x – 5) and (x + 3). If we consider some specific values of x as an input to f , we see that the output value f (x) can be determined by evaluating each factor, and then multiplying the factors to find the corresponding value of f (x).

ZEROS (ROOTS) OF QUADRATIC FUNCTIONS

a. Evaluate f at each of these input values by evaluating each factor, and then multiplying the values of the factors:

i) Given x = 6, evaluate f (6). ii) Given x = –1, evaluate f (–1).

i) Given x = 6, evaluate f (6).

ii) Given x = –1, evaluate f (–1).

iii) Given x = –0.5, evaluate f (–0.5).

We can use the distributive property to rewrite f (x) = (2x – 5)(x + 3) in the form f (x) = ax2 + bx + c. (This form is call standard form for a quadratic function.) As an example, f can be written as f (x) = 2x2 + x – 15 by expanding (2x – 5)(x + 3).

We expand a product of in this form by applying the distributive property. (Remember that the distributive property says that either a(b + c) = ab + ac or (a + b)c = ac + bc.)

b. Use the standard form of f (that is, f (x) = 2x2 + x – 15) to evaluate the function at the given input values:

i) Given x = 6, evaluate f (6). ii) Given x = –1, evaluate f (–1). iii) Given x = –0.5, evaluate f (–0.5).

i) Given x = 6, evaluate f (6).

ii) Given x = –1, evaluate f (–1).

iii) Given x = –0.5, evaluate f (–0.5).

c. What observations can you make from comparing how the values of f (6), f (–1), f (0.5) compare in questions (a) and (b) above?

The corresponding values in parts (a) and (b) are identical.

In Exercises #2-7 you are given quadratic functions written in factored form. Determine the standard form for each function by expanding the product of factors.

2. f (x) = (x + 1)(x + 7) 3. g(x) = (x – 2)(x – 5)

4. h(x) = (x + 3)(x – 10) 5. f (x) = (2x – 6)(x + 1)

6. p(x) = –3(x + 2)(5x + 11) 7. f (x) = 6(x – 4)2

2. f (x) = (x + 1)(x + 7)

3. g(x) = (x – 2)(x – 5)

4. h(x) = (x + 3)(x – 10)

5. f (x) = (2x – 6)(x + 1)

6. p(x) = –3(x + 2)(5x + 11)

7. f (x) = 6(x – 4)2

8. Recall the washer problem from Worksheet 3: A company makes bolts, screws, and washers. The washers are made with holes of different sizes to fit the various size bolts and screws according the following diagram. (For this exercise, recall that the area of a circle A is given by A = πr2 where r is the circle’s radius.)

a. The formula f (x) = π(x + 0.8)2 – π(x)2 calculates the area (in square cm) of the washer’s top surface if the radius of the circular hole is x cm. Expand the formula and simplify as much as possible.

b. What type of function is the relationship between the radius of the circular hole (in cm) and the area of the washer’s top surface (in square cm)?

The function is linear.

The zeros of a function are the input values such that the output of the function is 0. If f (x) = (2x – 5)(x + 3), then the zeros of f are the values of x such that f (x) = 0. (Note: Zeros are also called “roots” – the terms are used interchangeably.)

At this point, we need to learn an important fact called the zero-product property.

9a. Write four pairs of number such that multiplying the 1st number in the pair by the second number in the pair results in an answer of 0.

Answers will vary. Some examples are 5 and 0, 4 and 0, and 0 and –1.

b. What do these pairs of numbers have in common?

They all have 0 as one of the numbers in the pair.

c. Suppose we have two numbers a and b, and suppose that the product of these numbers is 0. That is, a∙b = 0. What does this tell us about the values of a and b?

Either a = 0 or b = 0.

Let’s return to function f. The zeros of f are the values of x such that f (x) = 0, or the values of x such that

(2x – 5)(x + 3) = 0

since f (x) = (2x – 5)(x + 3).

Note that the statement (2x – 5)(x + 3) = 0 says that we have two numbers, namely (2x – 5) and (x + 3), such that their product is 0.

This tells us that at least one of these numbers must be 0.

Therefore, either 2x – 5 = 0 or x + 3 = 0 must be true.

2x – 5 = 0 2x = 5 x = 2.5

x + 3 = 0 x = –3

Either 2x – 5 = 0 or x + 3 = 0 must be true.

The zeros of f are x = 2.5 and x = –3.

10. Evaluate f (2.5) and f (–3) to verify that they are zeros (roots) of f. (Recall f (x) = (2x – 5)(x + 3).)

When the output of a function is 0 for a real number input, then the graph crosses the horizontal axis at that input value.

In other words, we know f has horizontal intercepts at (2.5, 0) and (–3, 0).

11. If the zeros of the function are real numbers, what feature of the function’s graph do the zeros represent? (In other words, what do we know about the graph of the function if f (2.5) = 0 and f (–3) = 0?)

The value equidistant from x = 2.5 and x = –3 is the average of 2.5 and –3.

x = –0.25 must be the axis of symmetry. Points with the same output value are equidistant from the axis of symmetry for a quadratic function, and (2.5, 0) and (–3 , 0) are equidistant from x = –0.25.

12. What value of x is equidistant from the zeros (roots) of f ? What does this x-value represent for the quadratic function? How do you know?

In Exercises #13-18, find the zeros of each function (Hint: use the zero-product property). [Note: These are the same functions you worked with in Exercises #2-7.]

13. f (x) = (x + 1)(x + 7) 14. g(x) = (x – 2)(x – 5)

15. h(x) = (x + 3)(x – 10) 16. f (x) = (2x – 6)(x + 1)

17. p(x) = –3(x + 2)(5x + 11) 18. f (x) = 6(x – 4)2

13. f (x) = (x + 1)(x + 7)

14. g(x) = (x – 2)(x – 5)

15. h(x) = (x + 3)(x – 10)

16. f (x) = (2x – 6)(x + 1)

17. p(x) = –3(x + 2)(5x + 11)

18. f (x) = 6(x – 4)2

19. What is the axis of symmetry for the functions in Exercises #13 and #14?The axis of symmetry is equidistant from the roots, so we find it by averaging the values of the roots.

For Exercise #12: For Exercise #13:

20. What are the coordinates of the vertex for each of the functions in Exercises #13 and #14?

We know that the vertex lies on the axis of symmetry. So we evaluate the functions for the input that matches the axis of symmetry.

For Exercise #12: For Exercise #13: The vertex for Exercise #12 is (–4, –9) and for Exercise #13 is (3.5, –2.25).

21. Sketch a graph of the functions in Exercises #13 and #14.

The graph for Exercise #12:

The graph for Exercise #13:

22. Recall that the vertical intercept of a function is the output of the function when the input is 0. That is, the vertical intercept of a function f is f (0).

a. Find the vertical intercept of f if f (x) = –2x2 – 6x + 5.

The vertical intercept is located at (0, 5).

b. Find the vertical intercept of f if f (x) = (2x – 3)(x + 4).

The vertical intercept is located at (0, –12).

23. A function is given in standard form as f (x) = –2x2 + 11x – 12 and in factored form as f (x) = (–2x + 3)(x – 4). Find the function’s roots, vertical intercept, and vertex, then graph the function.

The standard form shows us that the vertical intercept occurs at (0, –12).

From the factored form it’s relatively easy to find that the roots are x = 1.5 and x = 4 (so the function has horizontal intercepts at (1.5, 0) and (4, 0)).

Averaging the roots gives us the line of symmetry, x = 2.75, and evaluating f (2.75) tells us that the vertex occurs at (2.75, 3.125).

24. We are given the graph of f, whose factored form looks like f (x) = a(x – u)(x – v).

a. What are the values of u and v for this function?

24. We are given the graph of f, whose factored form looks like f (x) = a(x – u)(x – v).

–4 and 1, so we can rewrite the function as f (x) = a(x – (–4))(x – 1)

or f (x) = a(x + 4)(x – 1).

f (x) = a(x + 4)(x – 1)b. Changing the value of a can create many different functions

with the same roots. To find the value of a that will produce the given graph we can substitute a known ordered pair for the values of x and y and solve the resulting equation for a. Given that f (–3) = –6, find the value of a that will generate the graph of f.

c. Write the formula for function f.

f (x) = 1.5(x + 4)(x – 1)

25. We are given the graph of g, whose factored form looks like g(x) = a(x – u)(x – v).

–1.5 and 2, so we can rewrite the function as g(x) = a(x – (–1.5))(x – 2)

or g(x) = a(x + 1.5)(x – 2).

25. We are given the graph of g, whose factored form looks like g(x) = a(x – u)(x – v).

g(x) = a(x + 1.5)(x – 2)

b. Given that g(1) = 5.25, find a value of a that will generate the graph of g.

c. Write the formula for function g.

g(x) = –2.1(x + 1.5)(x – 2)

d. Why can’t we use one of the horizontal intercepts in part (b) to find the value of a? For example, why can’t we use the fact that g(2) = 0 to find the value of a?

Regardless of the value of a, this statement is true. Therefore, it doesn’t help us determine the value for a.

Suppose we did. Then

1. What are the advantage(s) of writing the function in standard form?

FACTORING QUADRATIC FUNCTIONS

In Worksheet #4 we learned about the factored form of a quadratic function (for example, f (x) = (x – 2)(x + 5) or g(x) = (–2x + 3)(x – 4)) and the standard form of a quadratic function (for example, f (x) = x2 + 3x – 10 or g(x) = –2x2 + 11x – 12).

Answers may vary. It is easy to determine the vertical intercept or the concavity in standard form. It’s also easy to recognize that the function is quadratic when it’s written in standard form.

2. What are the advantage(s) of writing the function in factored form?

Since there is an advantage to writing a quadratic function in factored form, it’s important that we know how to factor quadratic expressions. The key to knowing how to factor is first to think about how to expand from factored form to standard form and imagine how to reverse this process.

Consider the function f (x) = (2x – 5)(x + 1). Let’s expand this to write the formula in standard form paying close attention to the steps in the process.

Answers may vary. Factored form makes it easy to determine the roots of the function.

If we instead began with f (x) = 2x2 – x – 10 and wanted to write the function in factored form, we would need to fill in the parentheses for f (x) = ( + )( + ) such that ( + )( + ) = 2x2 – x – 10.

To make it easier to reference, let’s write this as (a + c)(b + d) = 2x2 – x – 10.

We need to find a, b, c, and d such that ab = 2x2, cd = –10, and ad + cb = –x.

• The best place to start is to look at the possible values of a and b. There aren’t too many options using integer coefficients. a and b must be 2x and x so that ab = 2x2.

So far, then, we have (2x + c)(x + d) = 2x2 – x – 10.

• Next, we observe that cd = –10, so c and d must be factors of –10. We have the following possibilities:o c = –1, d = 10 or c = 10, d = –1o c = –10, d = 1 or c = 1, d = –10o c = –5, d = 2 or c = 2, d = –5o c = –2, d = 5 or c = 5, d = –2

• Since ad + cb = –x, we need to figure out which combination from the previous information gives us the correct factors.

o If c = –1 and d = 10, then ad + cb = (2x)(10) + (–1)(x) = 20x – x

= 19x ≠ –x

o If c = 10 and d = –1, then ad + cb = (2x)(–1) + (10)(x) = –2x + 10x = 8x ≠ –x

• Since ad + cb = –x, we need to figure out which combination from the previous information gives us the correct factors.

o If c = 1 and d = –10, then ad + cb = (2x)(–10) + (1)(x) = –20x + x = –19x ≠ –x

o If c = –10 and d = 1, then ad + cb = (2x)(1) + (–10)(x) = 2x – 10x = –8x ≠ –x

o If c = –5 and d = 2, then ad + cb = (2x)(2) + (–5)(x) = 4x – 5x = –x *

We can stop when we’ve found the right combination.

So the factored form of f (x) = 2x2 – x – 10 is

f (x) = (a + c)(b + d) = (2x + (–5))(x + 2) = (2x – 5)(x + 2).

Until you get proficient with factoring, it’s highly recommended that you expand the factored form you found to make sure that it is equivalent to the original formula.

Let’s look at another example. Let f (x) = x2 – 3x – 28. We want to write this in factored form f (x) = (a + c)(b + d) such that (a + c)(b + d) = x2 – 3x – 28.

• a = x and b = x since ab = x2. So (x + c)(x + d) = x2 – 3x – 28.

• We know that c and d are factors of –28. We have a lot of possibilities:

o c = –4, d = 7o c = –7, d = 4o c = –1, d = 28o c = –28, d = 1o c = –2, d = 14o and so on…

• Since ad + cb = –3x, we need to figure out which combination from the previous information gives us the correct factors.

o If c = –4 and d = 7, then ad + cb = (x)(–4) + (7)(x) = –4x + 7x

= 3x ≠ –3x

o If c = –7 and d = 4, then ad + cb = (x)(–7) + (4)(x) = –7x + 4x = –3x *

The factored form of f (x) = x2 – 3x – 28 is f (x) = (x – 7)(x + 4). Let’s check our answer by expanding the factored form.

A few more abbreviated examples follow.

If f (x) = x2 + 7x + 6, then f (x) = (x + 1)(x + 6) since •(x)(x) = x2 •(1)(6) = 6, and •(x)(6) + (1)(x) = 6x + x = 7x

If f (x) = x2 + 6x + 9, then f (x) = (x + 3)(x + 3) (also written f (x) = (x + 3)2) since •(x)(x) = x2, •(3)(3) = 9, and •(x)(3) + (3)(x) = 3x + 3x = 6x

If f (x) = 3x2 – 19x – 14, then f (x) = (3x + 2)(x – 7) since •(3x)(x) = 3x2, •(2)(–7) = –14, and •(3x)(–7) + (2)(x) = –21x + 2x = –19x

If f (x) = x2 – 16, we observe that f (x) = x2 + 0x – 16, and then f (x) = (x + 4)(x – 4) since •(x)(x) = x2, •(4)(–4) = –16, and •(x)(–4) + (4)(x) = –4x + 4x = 0x

In Exercises #1-10, write the factored form of the quadratic function.

1. f (x) = x2 + 7x + 12 2. g(x) = x2 + 5x – 14

3. h(x) = x2 – 5x – 14 4. g(x) = x2 – 3x + 2

5. f (x) = 2x2 + 9x + 7 6. h(x) = –x2 + 7x – 10

7. f (x) = 2x2 + x – 15 8. a(x) = 6x2 – 7x – 5

9. h(x) = 49x2 – 81 10. m(x) = 25x2 – 10x + 1

f (x) = (x + 4)(x + 3) g(x) = (x + 7)(x – 2)

h(x) = (x – 7)(x + 2) g(x) = (x – 2)(x – 1)

f (x) = (2x + 7)(x + 1) h(x) = –(x – 2)(x – 5)

f (x) = (2x – 5)(x + 3) a(x) = (2x + 1)(3x – 5)

h(x) = (7x – 9)(7x + 9) f (x) = (5x – 1)2

In Exercises #11-16 you are given quadratic functions written in standard form. For each, do the following.a) Give the coordinates of the function’s vertical intercept.b) Write the function in factored form.c) State the zeros of the function.d) Find the axis of symmetry.e) Give the coordinates of the vertex.

11. f (x) = x2 + 5x + 6 12. f (x) = x2 – 4x – 21

13. f (x) = –x2 + 6x – 8 14. f (x) = 2x2 – 18x + 40

15. f (x) = 3x2 + 6x – 9 16. f (x) = 4x2 – 9

11. f (x) = x2 + 5x + 6

a. (0, 6)b. f (x) = (x + 3)(x + 2)c. x = –3 and x = –2d. x = –2.5e. f (–2.5) = –0.25, so the vertex is (–2.5, –0.25)

12. f (x) = x2 – 4x – 21

a. (0, –21)b. f (x) = (x – 7)(x + 3)c. x = 7 and x = –3d. x = 2e. f (2) = –25, so the vertex is (2, –25)

13. f (x) = –x2 + 6x – 8

a. (0, –8)b. f (x) = –(x – 4)(x – 2)c. x = 4 and x = 2d. x = 3e. f (3) = 1, so the vertex is (3, 1)

14. f (x) = 2x2 – 18x + 40

a. (0, 40)b. f (x) = 2(x – 5)(x – 4)c. x = 5 and x = 4d. x = 4.5e. f (4.5) = –0.5, so the vertex is (4.5, –0.5)

15. f (x) = 3x2 + 6x – 9

a. (0, –9)b. f (x) = 3(x + 3)(x – 1)c. x = –3 and x = 1d. x = –1e. f (–1) = –12, so the vertex is (–1, –12)

16. f (x) = 4x2 – 9

a. (0, –9)b. f (x) = (2x – 3)(2x + 3)c. x = 1.5 and x = –1.5d. x = 0e. (0, –9)

17. You are given the graph of the function f (x) = 2x2 – 7x – 3. What does it mean to solve the equation 2x2 – 7x – 3 = 12? Explain, then represent the meaning on the graph.

Solving the equation 2x2 – 7x – 3 = 12 for x means that we seek the value(s) of the input value x such that the function f (x) = 2x2 – 7x – 3 outputs 12.

We can see this on the graph by looking for the x-value of the intersection points of y = 2x2 – 7x – 3 and y = 12.

Factoring and using the zero-product property is useful for solving quadratic equations in addition to finding the zeros of the function.

The solutions appear to be x = –1.5 and x = 5.

We can attempt to solve the equation using factoring. To begin with, we rewrite the equation so that the value of each side of the equation is 0. Then we factor and use the zero-product property.

18. The solutions to the equation 2x2 – 7x – 3 = 12 are x = –1.5 and x = 5. What do these values represent?

x = –1.5 and x = 5 are the input values for f (x) = 2x2 – 7x – 3 that generate an output value of 12

In Exercises #19-22, use factoring and the zero-product property to solve the given equation, then explain what your solution represents.

19. x2 + 12x + 13 = –22 20. 8 = x2 – 13x + 30

21. 2x2 + 6x – 6 = 2 22. 6x2 + 5x – 9 = –10

19. x2 + 12x + 13 = –22

x = –5 and x = –7 are the input values for f (x) = x2 + 12x + 13 that generate an output value of –22

20. 8 = x2 – 13x + 30

x = 11 and x = 2 are the input values for f (x) = x2 – 13x + 30 that generate an output value of 8.

21. 2x2 + 6x – 6 = 2

x = 1 and x = –4 are the input values for f (x) = 2x2 + 6x – 6 that generate an output value of 2.

22. 6x2 + 5x – 9 = –10

x = –1/3 and x = –2 are the input values for f (x) = 6x2 + 5x – 9 that generate an output value of –10.

23. Suppose the functions f (x) = x2 – 3x – 25 and g(x) = 3x + 15 are graphed on the same coordinate plane. Where will the graphs intersect? (Solve algebraically first, then you may check your answer by graphing the functions on a calculator.)

We want the value(s) of x such that f (x) = g(x).

The graphs intersect at (10, 45) and (–4, 3).

24. Suppose the functions f (x) = 2x2 – 4x – 15 and g(x) = x2 – 3x + 5 are graphed on the same coordinate plane. Where will the graphs intersect? (Solve algebraically first, then you may check your answer by graphing the functions on a calculator.)

We want the value(s) of x such that f (x) = g(x).

The graphs intersect at (5, 15) and (–4, 33).

Not all quadratic expressions are factorable using integers. When this is the case, factoring becomes much more difficult and impractical.

For example, when attempting to write the function f (x) = x2 + 0.5x – 3 in the factored form f (x) = (a + c)(b + d), we know a = x and b = x since ab = x2. We also know that c and d must be factors of – 3.

When looking at integer possibilities we have c = 3, d = –1 (or vice versa) or c = –3, d = –1 (or vice versa) and we find that none of these options work. We are now left to examine countless non-integer factors of –3 to determine which factors of –3 sum to 0.5. When this situation arises, we generally use an alternative method that you will learn about later in the module.

In Exercises #25-30, rewrite the expression in factored form if possible. If it’s not possible, say “not factorable using integers”.

25. f (x) = x2 + 3x + 5 26. f (x) = x2 – 7x + 10

27. f (x) = x2 + 10x + 15 28. f (x) = x2 – x + 1

29. f (x) = 2x2 + x – 1 30. f (x) = 3x2 – 2x + 4

not factorable using integers

f (x) = (x – 5)(x – 2)

f (x) = (2x – 1)(x + 1)

Copyright © 2012 Carlson, O’Bryan & Joyner Worksheet #1: Introduction to Quadratic Functions...

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4.3–$Transformations$of$Parabolas$Worksheet$#1$ MPM2D% … worksheet -1.pdf · 4.3–$Transformations$of$Parabolas$Worksheet$#1$ MPM2D% Jensen% % 1.%Sketch%the%graphs%of%these%three%quadratic%relations%on%the%same%set%of%axes.%%

WORKSHEET SOLVE A QUADRATIC EQUATION BY COMPLETING … · WORKSHEET – SOLVE A QUADRATIC EQUATION BY COMPLETING THE SQUARE 3 =1±2 Add 1 to both sides. t = 1 + 2 or t = 1 – 2 Split

Lesson Solving Quadratic Equations Geometrically · Lesson: Solving Quadratics Geometrically . 1 Log-on to: & Download the Worksheet 2 Learn through the Video lectures & Try Solving

Understanding Quadratic Functions and Solving Quadratic ...

8-5 Solving Quadratic Equations by Graphing · Solving Quadratic Equations by Graphing Objective Solve quadratic equations by graphing. Vocabulary quadratic equation Every quadratic

Downloaded from ...Title CBSE Class 11 Mathematics Worksheet - Complex Numbers and Quadratic Equation (4).pdf Author Subject CBSE Class 11 Mathematics Worksheet Complex Numbers and

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