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

2. 10-2: Quadratic Functions In 10-1, we graphed two types of equations y = ax 2, 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 = ax 2 + 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.

3. 10-2: Quadratic Functions In 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 “x 2 ” (that’s “a”), and plug them into the equation above. This is the value where you should start your table of values

4. 10-2: Quadratic Functions Example 1: Graphing y = ax 2 + bx + c Make a table of values and graph the quadratic function y = -3x 2 + 6x + 5 Step 1: Find the vertex -b / 2a = -(6) / 2(-3) = -6 / -6 = 1 Step 2: Make a table of values xy = -3x 2 + 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)

5. 10-2: Quadratic Functions Y OUR T URN Make a table of values and graph the quadratic function y = x 2 – 6x + 9 Step 1: Find the vertex -b / 2a = -(-6) / 2(1) = 6 / 2 = 3 Step 2: Make a table of values xy = x 2 – 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)

6. 10-2: Quadratic Functions Remember that the vertex represents the maximum or minimum value of a function. Example 2 The equation h = -16t 2 + 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

7. 10-2: Quadratic Functions Y OUR T URN 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 function h = -16t 2 + 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

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