essential question: how do you find the vertex of a quadratic function?

Essential Question: How do you find the vertex of a quadratic function?

10-2: Quadratic FunctionsIn 10-1, we graphed two types of equations

y = ax2, where the axis of symmetry = 0 and vertex was always (0,0)

Including c in the equation didn’t change the axis of symmetry (left/right), but did change the vertex (up/down)y = ax2 + c, where the axis of symmetry = 0

and vertex was at (0, c)Including b in the equation changes the axis

of symmetry and vertex.

10-2: Quadratic FunctionsIn standard form, the axis of symmetry is

found by using the equation:

Take the number in front of the “x” (that’s “b”) and the number in front of the “x2” (that’s “a”), and plug them into the equation above.

This is the value where you should start your table of values

2

bx

a

10-2: Quadratic FunctionsExample 1: Graphing y = ax2 + bx + c

Make a table of values and graph the quadratic functiony = -3x2 + 6x + 5

Step 1: Find the vertex -b/2a = -(6)/2(-3) = -6/-6 = 1

Step 2: Make a table of valuesx y = -3x2 + 6x + 5

(x, y)

1 -3(1)2 + 6(1) + 5 = 8

(1, 8)

2 -3(2)2 + 6(2) + 5 = 5

(2, 5)

3 -3(3)2 + 6(3) + 5 = -4

(3, -4)

2 4 6 8 10–2–4–6–8–10 x

2

4

6

8

10

–2

–4

–6

–8

–10

y

10-2: Quadratic FunctionsYOUR TURN

Make a table of values and graph the quadratic functiony = x2 – 6x + 9

Step 1: Find the vertex -b/2a = -(-6)/2(1) = 6/2 = 3

Step 2: Make a table of valuesx y = x2 – 6x + 9 (x, y)

3 (3)2 – 6(3) + 9 = 0

(3, 0)

4 (4)2 – 6(4) + 9 = 1

(4, 1)

5 (5)2 – 6(5) + 9 = 4

(5, 4)

10-2: Quadratic FunctionsRemember that the vertex represents the

maximum or minimum value of a function.Example 2

The equation h = -16t2 + 72t + 520 gives the height in feet h of a firework shot into the air after t seconds. How long will it take for the firework to reach its maximum height? At that time, how far above the ground will it be?

Use -b/2a to find the vertex -(72)/2(-16) = -72/-32 = 2.25 seconds

Substitute 2.25 into the equation -16(2.25)2 + 72(2.25) + 520 = 601 ft

10-2: Quadratic FunctionsYOUR TURN

A ball is thrown into the air with a velocity of 48 ft/s. Its height h in feet after t seconds is given by the functionh = -16t2 + 48t + 4.

In how many seconds will the ball reach its maximum height? -(48)/2(-16) = -48/-32 = 1.5 seconds

What is the ball’s maximum height? -16(1.5)2 + 48(1.5) + 4 = 40 ft

10-2: Quadratic FunctionsAssignment

Worksheet #10-2Problems 1 – 21 (odds), 31 & 33