# How do I find the x-intercepts of a quadratic function in vertex form f(x)=1/4(x+2)^2-9?

May 9, 2015

The x-intercepts of an equation happen at those points when
$f \left(x\right) = 0$

Given $f \left(x\right) = \frac{1}{4} {\left(x + 2\right)}^{2} - 9$
If $f \left(x\right) = 0$

$\frac{1}{4} {\left(x + 2\right)}^{2} - 9 = 0$

$\frac{1}{4} {\left(x + 2\right)}^{2} = 9$

${\left(x + 2\right)}^{2} = 36$

$\left(x + 2\right) = \pm 6$

$x = 4$ or $x = - 8$

May 9, 2015

Develop to standard form:
$f \left(x\right) = \frac{1}{4} \left({x}^{2} + 4 x + 2\right) - 8 = {x}^{2} / 4 + x + 1 - 9 = {x}^{2} + 4 x - 32 = 0$

Find 2 numbers knowing their sum( -b = -4) and their product (c = -32). Compose factor pairs of (-32). Proceed: (-1, 32)(-2, 16)(-4, 8). This last sum is 8 - 4 = 4 = b. Then, the 2 real roots, or x-intercepts are the opposite: 4 and -8.
Answers: x-intercepts : 4 and -8.
Check:
x = 4 -> $f \left(4\right) = \frac{36}{4} - 9 = 9 - 9 = 0$ Correct
x = -8 ->$f \left(- 8\right) = \frac{36}{4} - 9 = 9 = 9 = 0$. Correct.