Today:

Review from Last Week Getting to Know the Quadratic Function:

(how the b value changes the parabola) Class Work

Grades:1. Class work: (20%) Ten graded assignments. How many did you

submit with all work showing? Answers only receive no credit for those problems.2. Home work: (20%) 15 Khan Academy Topics. How many did you complete?3.Tests: (40%) Six tests, lowest grade dropped. All test scores are posted online. You can find your average test score.

4. Final Exam: (10%) This quarter's final could not hurt your average. Everybody wins here.5. Notebook: (10%) Is it Complete? Warm up problems? Class notes? Is it organized? Sections labeled? Is it easy to read?Is it a notebook full of doodles, notes to friends and folded, not

submitted class work?Warm-Up Section of Notebook

Find the axis of symmetry for the graph y = 3x2 + 6x + 4

a. y = -1 b. x = 1 c. x = -1 d. y = 1

A graph of a quadratic function has x intercepts of (3,0) and (-7,0). Which of the following could match the graph?

a. x2 + 4x - 21 = 0 b. x2 - 10x - 21 = 0 c. x2 + 10x + 21 = 0 d. x2 - 4x + 21 = 0

Warm-Up/Review:

Warm-Up/Review:

1. y = 5x2 – 1; Graph the general shape and location from the information given in the function. Width, shape, AOS, y-intercept, etc. Even the missing ‘b’ term provides information.

Warm-Up/Review:

Warm-Up/Review:

Solve and graph the function: -x2 + 2x = 1

Class Notes Section of Notebook

Important Concepts to Review:1. There is no single "right way" to graph a quadratic function.

In fact the goal is for you to understand and use a variety of methods, so that you can choose the best (easiest) method for a given problem.

2. The axis of symmetry is an important part of parabolas and can save you time and effort if you understand its properties.

Because a parabola is symmetrical, two points opposite each other on the curve are the same number of units away from the axis of symmetry.

Helpful Hint

2.5 You must have at least 5 points to graph the parabola.

4. The vertex is an (x, y) coordinate on the AOS, and is either theminimum or maximum y value5. All parabolas begin from the parent function (y = x2), and are moved around the coordinate plane from changes in the a, b, and c values.

3. The axis of symmetry has a single coordinate (x) and represents the exact center of the parabola.

6. A quadratic equation with no solutions will not cross the x-axis at any point. It can still be graphed, however.

Important Concepts to Review:

How the a and c values affect the quadratic function y = ax2 + bx + c

Start with the parent function, which is...First, how does a change in a affect the parabola

y = x2

Review: Effects of the a, b, & c values

Effects of the a, b, & c values

1. The greater the value of 'a', the narrower and steeper the graph.2. A positive 'a' value results in parabola which turns up and has a vertex minimum.

3. A negative 'a' value results in parabola which turns down and has a vertex maximum.

Effects of the a, b, & c values

How does a change in c affect the parabola?

The value of c is also used to find the y-intercept.

Set the 'x' values = 0, and find the intercept.

Can we determine the equation of this graph?

How changes in 'b' affect the parabola:

Why does a positive b value (see aqua, b = 2) result in a shift 2 units

to the left?

Effects of the a, b, & c values

Name five important parts of a parabola

1. Axis of Symmetry2. Vertex3. y-intercept4. y-intercept translated5. Solution(s)

How do we find each of these?

Axis of symmetry: Use the formula: -b2aVertex:

The AOS is the x-coordinate of the vertex. To find the y-coordinate, plug the value of x into the equation and solve for y.

y-intercept

Write a quadratic equation in standard form: ax2 + bx + c = 0

The y-intercept is the value of 'c' in the quadratic equation.

y-intercept translated

Determine the distance of the y axis from the AOS. Then, count the same distance on the other side of the AOS. The y value will be the same.

Symmetry: 5. Solution(s)

The solutions are where the graph crosses the x-intercept. Right now, our method for finding solutions is by factoring.

We will be learning 3 more methods of finding solutions:

1. By using square roots.2. By completing the square3. by using the quadratic function

Graph the quadratic function. y = x2 + 4x + 2

Step 2 Find the axis of symmetry,

Step 1: Try to picture what the graph will look like before you start. Use the a, b, and c values to determine your prediction

Step 3: Determine the best method(s) to solve that particular function.

Step 4 : Plot at least 5 points, then connect the dots to complete the parabola.

then find 'y' to complete the coordinates for the vertex

Graphing Quadratic Functions: (3)

Step 2 Find the axis of symmetry and the vertex.

Substitute for x to find the y coordinate

The x-coordinate of the vertex is...

The y-coordinate is Find at least 4 more points, then graph.

This is also the AOS

y = x2 + 4x + 2

Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points.

Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve.

y = x2 + 4x + 2

STEP 1 Identify the coefficients of the function. STEP 2 Find the vertex. Calculate the x - coordinate.STEP 3 Draw the axis of symmetrySTEP 4 Identify the y - intercept c, STEP 5 Find the roots by using one of the solution methods,

We are unable to find the roots with our knowledge for now, so we'll select another value of x and solve for y. The AOS is 1, so let's choose x = -1. Find the y coordinate.

The two other points are (–1, 10) and (–2, 25) STEP 6 Reflect this point over the AOS to plot another point.

STEP 7 Graph the parabola

Graph a function of the form y = ax2 + bx + cy = 3x2 – 6x + 1, Plot 5 points and draw the curveGraph

x = 1

(–1, 10)

(0, 1)

(1, –2)

(–2, 25)

Graph a function of the form y = ax2 + bx + c

Step 1: Find the axis of symmetry.Use x = . Substitute 1 for a

and –6 for b.

The axis of symmetry is x = 3.

= 3

y = x 2 – 6x + 9 Rewrite in standard form.y + 6x = x2 + 9Graph the quadratic function

Step 2: Find the vertex.

Simplify.= 9 – 18 + 9

= 0

The vertex is (3, 0).

The x-coordinate of the vertex is 3. Substitute 3 for x.

The y-coordinate is 0.

y = x2 – 6x + 9

y = 32 – 6(3) + 9

Graph a function of the form y = ax2 + bx + cy = x 2 – 6x + 9

Step 3: Find the y-intercept.y = x2 – 6x + 9

y = x2 – 6x + 9

The y-intercept is 9; the graph passes through (0, 9).

Identify c.

Graph a function of the form y = ax2 + bx + cy = x 2 – 6x + 9

Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y- intercept.Since the axis of symmetry is x = 3, choose x-values less than 3.

Let x = 2y = 1(2)2 – 6(2) + 9

= 4 – 12 + 9= 1

Let x = 1 y = 1(1)2 – 6(1) + 9

= 1 – 6 + 9= 4

Substitutex-coordinates.

Simplify.

Two other points are (2, 1) and (1, 4).

Graph a function of the form y = ax2 + bx + cy = x 2 – 6x + 9

Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points.

Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve.

y = x 2 – 6x + 9

x = 3

(3, 0)

(0, 9)

(2, 1)(1, 4)

(6, 9)

(5, 4)

(4, 1)

x = 3

(3, 0)

Class Work 4.1~You can skip problem 2 on the back. Every other problem is required.Be sure to show all your work for credit.

Class Work 4.2~Find the important points on the graph. Label them, and draw the curve.Show your work.