Chapter 6Quadratic Functions

● Determine the characteristics of quadratic functions

● Sketch quadratic functions● Solve problems modelled by quadratic functions

6.1: Quadratic Functions● A quadratic function is of the form: y = ax2 + bx + c

– Where a, b, c are real number constants

– Also

– Example: y = 2x2 + 3x – 9, y = x2

● The graph of a quadratic function has a distinct shaped called a parabola

● The parabola can either either “open upwards” or “open downwards”

Uses of Quadratics and Parabolas

Projectile Motion

● Anything that is thrown that has some horizontal motion.

– Jumping on a bike, skis, snowboards, skidoos, etc.

– Running off a diving board.

– Arrows, or bullets that are shot.

– Throwing footballs or baseballs, or kicking a soccer ball.

● Parabolic skis

● When you take a slap shot, you stick flexes and forms a parabola

● The concave on a skateboard creates a parabola

● A skateboard ramp is a parabola

● Accelerated motion in science 1206– d = vt + 0.5at2

Parts of a Parabola

Parts of a Parabola

Quadratic Functions and Parabolas

● Any relation that can be represented by a parabola can be modelled by a quadratic function.

● A quadratic function must have a quadratic (or a squared) term has its highest degree.

● The general equation for a quadratic is:– y = ax2 + bx + c

● This is called the Standard Form● ax2 is the quadratic or squared term● bx is the linear term● c is the constant term

Examples

Which of the following are quadratic functions?

If not, why is it not?

If it is, what direction does it open?

1) y = 3x + 1 not quadratic b/c no x2 term

2) y = 2x2 + x + 1 opens upward

3) y = -2x2 + x3 + x not quadratic b/c x3 term

4) y = 6 – 5x2 opens downward

5) y = (x+1)(x-3) = x2 – 3x + x – 3 = x2 -2x -3– Opens upward

6) y = (x+2)2 = x2 + 4x +4 opens upward

7) y = (x-3)2 + 1 = x2 -6x +10 opens upward

Putting disguised quadratics into general form (ax2 + bx + c = 0)

So why do we need to be able to put quadratics into general form?

● So we can determine the values of a, b, and c. ● These values tell us about the shape of the

parabola. ● For example, what does “a” tell us?

– It tells us whether the parabola is opened up or down. ● What else do these values tell us?

– Lets find out!

Effects of “a” on the parabola

● First investigate the effect of changing the value of a 1. What happens to the direction of the opening of the quadratic if a < 0 and a > 0? Opens downward Opens upward2. A) If the quadratic opens upward, is the vertex a maximum or minimum point? MinimumB) What if the quadratic opens downward? Maximum

3. Is the shape of the parabola affected by the parameter a ?

In other words are some graphs wider or narrower?

◦Yes! Larger values of “a” (in both the negative and positive direction) make the parabola narrower.

◦Values of “a” close to zero make the graph wider.

4. What happens to the x-intercepts as the value of a is changed? Change in value or cease to exist5. What is the impact on the graph if a = 0? Not a parabola

Effects of “b” on the parabola

6. What is the effect of parameter b in

y = ax2 + bx + c?

◦Changing the value of b changes the location of the vertex

7. Is the parabola’s line of symmetry changing?

◦Yes! It changes with the vertex

Effects of “c” on the parabola8. What is the effect of parameter c in

y =ax2 +bx + c?

◦Changing the value of c changes the location of the y-intercept. It moves the graph up or down

9. How can you identify the y-intercept from the equation in general form?

◦The value of c is the y-intercept.

10. Is the line of symmetry affected by the parameter c?

◦No! It does change. The x-value of the vertex is the same.

Sketching Parabolas

Another way of finding the vertex● Consider the last problem: y = 3x2 – 6x + 10● What are the values of:

– a = 3 b = -6 c = 10

● Calculate:

● How does this value relate to the vertex?

– It gives the x-coordinate of the vertex!

● The y-value of the vertex is found by putting the x-value into the quadratic function

Find the vertex of the following

Find the maximum or minimum y-value for the following

Find the range and the axis of symmetry for the following

● The height of a soccer ball kicked into the air is given by the function y = -5x2 + 20x. Determine the maximum height of the ball and the time that it occurred.

Finding the vertex from the average of x-values

● The x-coordinate of the vertex (the axis of symmetry) can also be found by taking average of the x-values for any two points on the parabola that has the same y-values.

Table of values

Given Points

6.3: Drawing more accurate parabolas

● A quick sketch of a parabola can be made if you know the vertex and the direction the parabola opens.

● Example: Sketch the graph of y = x2 +2x-3

● The sketch the graph of y = x2 +2x-3 can be made more accurate if we know more points.

● The extra points that are typically used are the places where the function crosses the x and y axis. – Also known as the x and y intercepts, respectively

● Y-intercept – To find the y-intercept we set x = 0 and solve for y – Y = 02 + 2(0) -3 = -3

● Remember that for the function y =ax2 +bx + c the y-intercept is c

● Lets sketch the parabola again using the y-intercept.

● There is another point that we can use that has the same y value as the y-intercept located the same distance from the axis of symmetry as the y-intercept

● X-intercept – To find the x-intercept we set y = 0 and solve for x

– 0 = x2 +2x-3 ● When finding the x-intercepts of a function you are actually

finding the “zeros” for the function.

– ie. You are finding the x-values that make the function equal to zero.

● To solve for x we will need to factor the quadratic. ● You will be required to use factoring methods developed in

Mathematics 1201 to determine the zeros.

● SO, how do we factor x2 +2x-3 = 0

– Product and Sum Method

– You must find two numbers such that the product of the two numbers equals “c”, the constant term.

– The sum of the two numbers must equal “b”, the coefficient on x.

So now what?

● To find the zeros from the factored form of the function we use the Zero Product Property

● The zero product property states that: – if the product of two real numbers is zero, – then one or both of the numbers must be zero.

● Find the zeros of (x + 3)(x – 1) =0 (done on previous slide)– The zeros (ie the x-intercepts) are x = -3 and

x = 1

Lets sketch the parabola again with the x-intercepts included.

Examples

For each of the following find

A) the vertex

B) the y-intercept

C) the x-intercepts

Then use this information to draw a graph of the function.

1). y = x2 + 5x + 4

y = x2 + 5x + 4

2). y = -2x2 -4x + 30

y = -2x2 -4x + 30

3). y= 3x2 – 13x + 4

y= 3x2 – 13x + 4

Vertex form of a Quadratic

● Vertex form: y = a(x-h)2 + k– “a” is the leading coefficient

● If a is positive --> parabola opens upward● If a is negative --> parabola opens downward

– The point (h,k) is the vertex of the parabola

● What is the vertex of the following?– 1. y = 2(x-1)2 + 3 (1,3)

– 2. y = 0.5(x+2)2 – 5 (-2, -5)

Sketch

● What is the vertex? (-1,-2)● Which way does the graph open?

– Upward b/c a = 0.5 > 0

Sketch y = 2(x-1)2 + 3

● What is the vertex? (1,3)● Which way does the graph open? Upward● What is the equation of the axis of symmetry

– X = 1

● What is the min/max y value? Minimum of y = 3● What is the domain and range

–

What is the equation's general form? (y=ax2+bx+c)

1201 factoring!!

Sketch the graph of the following by finding the intercepts

Note

● Those last two equations were in a special form known as Factored Form of the quadratic

Factored form of the quadratic

● y = a(x-r)(x-s)– r and s are the zeros of the function

– (r,0) & (s,0) are the x-intercepts of the parabola.

– a is the leading coefficient

● When given the x-intercepts of a quadratic function, one can easily find the value of a and thus the specific equation in factored form if given a point (other than the x-intercepts that is on the parabola)

Example

● y = a(x+3)(x-4)– Take the point (3,-3)

● -3 = a(3+3)(3-4)● -3 = a(6)(-1)● -3 = -6a● A = -0.5● Thus: y = -0.5(x+3)(x-4)

WORD PROBLEMS!!