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Review Khan Academy Topics Graphing Quadratic Functions:

Using Square Roots to Solve Equations Class Work 4.2

STEP 1 Identify the coefficients of the function. STEP 2

Find the vertex. Calculate the x - coordinate.STEP 3Draw the axis of symmetry

STEP 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.

= 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)

Graphing Quadratic FunctionsAfter a player takes a shot, the height in feet of a basketball can be modeled by f(x) = –16x2 + 32x, where 'x ' is the time in seconds after it is shot. Find1. The basketball’s maximum height 2. The time it takes the basketball to reach this height. 3. How long the basketball is in the air.

There is no c term in this equation. What does that tell us about our graph??

The graph is not shifted up or down the y axis, therefore the y-intercept is at the origin, which also means one of the solutions must be zero.

Quadratic Applications:

Write, do not say the answer, please.

Graphing Quadratic Functions1 Understand the Problem

Our answer includes three parts:

1. The maximum height of the ball,

2. The time to reach the maximum height, and

3. The time to reach the ground.

• The function f(x) = –16x2 + 32x models the height of the basketball after x seconds.

List the important information:

What are the two variables for our x and y axes. (Plural of axis, pronounced ax-eez)

Graphing Quadratic Functions2 Make a Plan

The basketball will hit the ground when its height is 0. Round to the nearest whole number if necessary.

What parts of the graph are important in solving our problem?

A. The vertex. Why?

A. Because the maximum height of the basketball and the time it takes to reach it are the coordinates of the vertex.

B. The zero's of the function because......

Graphing Quadratic FunctionsSolve3

Step 1 Find the axis of symmetry.

Use x = . Substitute

–16 for a and 32 for b.

Simplify.

The axis of symmetry is x = 1.

Graphing Quadratic FunctionsStep 2 Find the vertex.

f(x) = –16x2 + 32x= –16(1)2 + 32(1)= –16(1) + 32= –16 + 32 = 16

The vertex is (1, 16).

The x-coordinate of the vertex is 1. Substitute 1 for x.

Simplify.The y-coordinate is 16.

Graphing Quadratic FunctionsStep 3 Find the y-intercept.

Identify c.f(x) = –16x2 + 32x + 0

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

Graphing Quadratic FunctionsStep 4: Graph the axis of symmetry, the vertex, and the point containing the y-intercept. Then use symmetry to reflect the point across the axis of symmetry. Connect the points with a smooth curve.

(0, 0)

(1, 16)

(2, 0)

Graphing Quadratic FunctionsThe vertex is (1, 16). So at 1 second, the basketball has reached its maximum height of 16 feet.

(0, 0)

(1, 16)

(2, 0)

The graph shows the zero’s of the function are 0 and 2. At 0 seconds the basketball has not yet been thrown, and at 2 seconds it reaches the ground. The basketball is in the air for 2 seconds.

Graphing Quadratic Functions

The vertex is the highest or lowest point on a parabola. Therefore, it always represents the maximum height of an object following a parabolic path.

Remember!

Tuesday's Class Work...Plus the following

Graphing Quadratic FunctionsClass Work 4.2

~Find the important points on the graph. Label them, and draw the curve.

Show your work.

Graphing Quadratic Functions

Graphing Quadratic Functions