1. 6.6 Analyzing Graphs of Quadratic Functions

5. Graphing the parabola y = f ( x ) = ax 2 + k Algebraic Approach: y = – 4 x 2 – 3 Numerical Approach: Graphical Approach: Consider the equation y = – 4 x 2 – 3 . What is a ? a = – 4 x Vertex (0, -3) -16 -4 0 -4 -16 -4 x 2 -19 -7 -3 -7 -19 - 4 x 2 - 3 2 1 0 – 1 – 2 x

6. y = – 4 x 2 – 3 . x Vertex (0, -3) y = – 4 x 2 In general the graph of y = ax 2 + k is the graph of y = ax 2 shifted vertically k units. If k > 0, the graph is shifted up. If k < 0, the graph is shifted down. (P. 267) The graph y = – 4 x 2 is shifted down 3 units.

7. a = – 4. What effect does the 3 have on the function? y x y = – 4 x 2 y = – 4( x – 3) 2 Consider the equation y = – 4( x – 3) 2 . What is a ? The axis of symmetry is x = 3. Numerical Approach: Axis of symmetry is shifted 3 units to the right and becomes x = 3 -16 -4 0 -4 -16 – 4 x 2 2 1 0 – 1 – 2 x -4 -16 -36 -64 -100 - 4 ( x- 3) 2 -36 3 0

16. Standard 9 Write a quadratic function in vertex form Write y = x 2 – 10 x + 22 in vertex form. Then identify the vertex. y = x 2 – 10 x + 22 Write original function. y + ? = ( x 2 – 10 x + ? ) + 22 Prepare to complete the square. y + 25 = ( x 2 – 10 x + 25 ) + 22 y + 25 = ( x – 5) 2 + 22 Write x 2 – 10 x + 25 as a binomial squared. y + 3 = ( x – 5) 2 Write in vertex form. Add – 10 2 2 ( ) = (–5) 2 = 25 to each side. The vertex form of the function is y + 3 = ( x – 5) 2 . The vertex is (5, – 3). ANSWER

17. EXAMPLE 1 Write a quadratic function in vertex form Write a quadratic function for the parabola shown. SOLUTION Use vertex form because the vertex is given. y – k = a ( x – h ) 2 Vertex form y = a ( x – 1) 2 – 2 Substitute 1 for h and –2 for k . Use the other given point, ( 3 , 2 ) , to find a . 2 = a ( 3 – 1) 2 – 2 Substitute 3 for x and 2 for y . 2 = 4 a – 2 Simplify coefficient of a . 1 = a Solve for a .

18. EXAMPLE 1 Write a quadratic function in vertex form A quadratic function for the parabola is y = ( x – 1) 2 – 2 . ANSWER

19. EXAMPLE 1 Graph a quadratic function in vertex form Graph y – 5 = – ( x + 2) 2 . SOLUTION STEP 1 STEP 2 Plot the vertex ( h , k ) = ( – 2, 5) and draw the axis of symmetry x = – 2. 14 Identify the constants a = – , h = – 2, and k = 5. Because a < 0, the parabola opens down. 14

20. EXAMPLE 1 Graph a quadratic function in vertex form STEP 3 Evaluate the function for two values of x . Plot the points (0, 4) and (2, 1) and their reflections in the axis of symmetry. STEP 4 Draw a parabola through the plotted points. x = 0 : y = ( 0 + 2) 2 + 5 = 4 14 – x = 2 : y = ( 2 + 2) 2 + 5 = 1 14 –

21. GUIDED PRACTICE for Examples 1 and 2 Graph the function. Label the vertex and axis of symmetry. 1. y = ( x + 2) 2 – 3 2. y = –( x + 1) 2 + 5

22. GUIDED PRACTICE for Examples 1 and 2 3. f ( x ) = ( x – 3) 2 – 4 12