1. Algebra 1
Section 12.8

2. Effect of Adding a Constant
f(x) = ax2 + c
If c is positive, the graph of f(x) = ax2 slides up c units.
If c is negative, the graph of f(x) = ax2 slides down c units.
The vertex is at (0, c).

3. f(x) = a(x – h)2 + k
The effect of adding a middle term, bx, is more easily seen when the equation is written in this form.
A constant h is subtracted from the independent variable before the expression is squared.

4. Example 1 To find the x-coordinate of the parabola’s vertex, solve:
Make a table of ordered pairs, including x = 3 and several points on either side of it.

5. Example 1 Make a table of ordered pairs. x (x – 3)2 g(x) 1
(1 – 3)2 = (-2)2 4 2
(2 – 3)2 = (-1)2 1 3
(3 – 3)2 = 02 4
(4 – 3)2 = 12 1 5
(5 – 3)2 = 22 4

6. Example 1
The graph of g(x) is the result of translating the graph of f(x) three units to the right.
Each point on g(x) is three units to the right of its corresponding point on f(x).

7. Example 2 To find the x-coordinate of the parabola’s vertex, solve:
Make a table of ordered pairs, including x = -1 and several points on either side of it.

8. Example 2 Make a table of ordered pairs. x (x + 1)2 h(x) -3
(-3 + 1)2 = (-2)2 4 -2
(-2 + 1)2 = (-1)2 1 -1
(-1 + 1)2 = 02
(0 + 1)2 = 12 1 1
(1 + 1)2 = 22 4

9. Example 2
The graph of h(x) is the result of translating the graph of f(x) one unit to the left.
Each point on h(x) is one unit to the left of its corresponding point on f(x).

10. Translations
The graph of f(x) = ax2 is translated |h| units horizontally when a constant h is subtracted from the variable before the expression is squared.

11. Translations
The graph of f(x) = a(x – h)2 slides to the right if h > 0 and slides to the left if h < 0.
The graph of f(x) = a(x – h)2 + k is the result of combining horizontal and vertical translations of f(x) = ax2.

12. Example 3 Graph p(x) = (x – 2)2 – 1.
The x-coordinate of the parabola’s vertex is 2.
The vertex is at (2, -1).
Compared to f(x) = x2, the graph has shifted 2 units to the right and 1 unit down.

13. f(x) = a(x – h)2 + k If x = h, f(h) = a(h – h)2 + k f(h) = a(0) + k
f(h) = k
Therefore, the vertex of the parabola is (h, k).

14. f(x) = a(x – h)2 + k Vertex of the parabola: (h, k).
The line of symmetry is a vertical line through the vertex, so its equation is x = h.

15. Example 4 Graph q(x) = -2(x – 3)2 + 5.
The parabola’s vertex is (3, 5).
The parabola’s line of symmetry is x = 3.

16. Example 4 Make a table of ordered pairs. x -2(x – 3)2 + 5 q(x) 1
-2(1 – 3)2 + 5 -3 2
-2(2 – 3)2 + 5 3 3
-2(3 – 3)2 + 5 5 4
-2(4 – 3)2 + 5 3 5
-2(5 – 3)2 + 5 -3

17. f(x) = a(x – h)2 + k
This form of a quadratic function is called the vertex form or the graphing form because it displays much information about the graph of a quadratic function.

18. f(x) = a(x – h)2 + k f(x) = -⅓(x + 7)2 + 9 f(x) = -⅓(x – (-7))2 + 9
Vertex: (-7, 9)
Line of symmetry: x = -7
Opens down, since a < 0
Flatter than f(x) = x2

19. f(x) = a(x – h)2 + k
If a quadratic function is in standard form, f(x) = ax2 + bx + c, it can be changed to the vertex form.

20. Opens upward; vertex at (3, 1)
Example 5
f(x) = x2 – 6x + 10
f(x) = (x2 – 6x) + 10
f(x) = (x2 – 6x + 9) + 10 – 9
f(x) = (x – 3)2 + 1
a = 1, h = 3, k = 1
Opens upward; vertex at (3, 1)

21. f(x) = a(x – h)2 + k It can be shown (p. 518) that... b h = - 2a b2
k = (c – ) b2 4a

22. Example 6 f(x) = -2x2 – 12x – 19 a = -2, b = -12 h = - -12 2(-2) = -4
= -3
k = f(-3)
= -2(-3)2 – 12(-3) – 19
= – 19
= -1

23. Example 6 h = -3 k = -1 The vertex is (-3, -1).
The line of symmetry is x = -3.
Since a = -2, the parabola opens downward.

24. Homework: pp