1. Algebra 1
Section 12.9

2. The y-intercept of a Function
Since the y-intercept always has an x value of zero, the y-intercept can be found by letting x = 0 and solving for y = f(0).

3. Example 1 g(x) = 2x2 – 3x + 5 g(0) = 2(0)2 – 3(0) + 5 = 5
y-intercept: (0, 5)
h(x) = (x – 1)2 + 2
h(0) = (0 – 1)2 + 2 = 3
y-intercept: (0, 3)

4. Zeroes of a Function
The x-intercepts of a graph always have y values of zero.
Therefore, the x-intercepts can be found by letting y = 0 or f(x) = 0 and solving for x.

5. Definition
A zero (or root ) of a function is a value of x at which the graph of the function crosses the x- axis. It is the x-coordinate of an x-intercept of the graph.

6. Example 2a Find the zero(s) of r(x) = x2 – 4x – 5 x2 – 4x – 5 = 0
x – 5 = 0 or x + 1 = 0
x = 5, -1

7. Example 2b s(x) = -x2 + 4x – 4 -x2 + 4x – 4 = 0 -1(x2 – 4x + 4) = 0
x – 2 = 0 or x – 2 = 0
x = 2 or x = 2

8. Double Root
In Example 2b, the single root is often called a double root since two factors produce the same root.
In this case, the parabola’s vertex is located at that x- intercept.

9. Example 3a Find the zero(s) of p(x) = (x – 1)2 – 6 (x – 1)2 – 6 = 0

10. Example 3a
x = 1 ± 6
x = ≈ 3.45
x = 1 – 6 ≈ -1.45

11. Example 3b Find the zero(s) of q(x) = x2 – x + 4 x2 – x + 4 = 0
1 ± -15 2
There are no real zeros.

12. No Real Zeros In Example 3b, there are no x- intercepts.
There are no x-intercepts or zeros when the discriminant is negative.

13. Example 4
f(x) = 3x2 + 8x – 3
The y-intercept is found by finding f(0).
f(0) = 3(0)2 + 8(0) – 3 = -3
y-intercept: (0, -3)

14. Example 4
f(x) = 3x2 + 8x – 3
The zeros are found by setting the function equal to zero.
3x2 + 8x – 3 = 0
(3x – 1)(x + 3) = 0
3x – 1 = 0 or x + 3 = 0

15. Example 4
f(x) = 3x2 + 8x – 3
The zeros are found by setting the function equal to zero.
3x – 1 = 0 or x + 3 = 0
x = ⅓ or x = -3
The zeros are ⅓ and -3.

16. Example 4
f(x) = 3x2 + 8x – 3
The vertex is (h, k). Use the formula to find h.
a = 3, b = 8, c = -3
h = - b 2a
= - 8
2(3)
= - 4 3
or -1⅓

17. Example 4 f(x) = 3x2 + 8x – 3 To find k, use k = f(h). k = f(- )
= 3(- )2 + 8(- ) – 3 4 3
= - 25 3
or -8⅓
The vertex is (- , ). 4 3 25
The vertex is (-1⅓, -8⅓).

18. Homework: pp