# How do you find the vertex and intercepts for y = (1/8)(x – 5)^2 - 3?

Apr 7, 2018

Vertex is at $\left(5 , - 3\right)$ y intercept is $y = \frac{1}{8}$ and
x intercepts are at ** $\left(5 + 2 \sqrt{6} , 0\right) \mathmr{and} \left(5 - 2 \sqrt{6} , 0\right)$

#### Explanation:

$y = \frac{1}{8} {\left(x - 5\right)}^{2} - 3$ Comparing with vertex form of

equation f(x) = a(x-h)^2+k ; (h,k) being vertex we find

here $h = 5 , k = - 3 \therefore$ Vertex is at $\left(5 , - 3\right)$, y intercept

is found by putting $x = 0$ in the equation $y = \frac{1}{8} {\left(0 - 5\right)}^{2} - 3$ or

$y = \frac{25}{8} - 3 \mathmr{and} y = \frac{1}{8} \therefore$ y intercept is $y = \frac{1}{8}$ or at

$\left(0 , \frac{1}{8}\right)$, x intercept is found by putting $y = 0$

in the equation $\therefore 0 = \frac{1}{8} {\left(x - 5\right)}^{2} - 3$ or

$\frac{1}{8} {\left(x - 5\right)}^{2} = 3 \mathmr{and} {\left(x - 5\right)}^{2} = 24 \mathmr{and} \left(x - 5\right) = \pm \sqrt{24}$ or

$x = 5 \pm 2 \sqrt{6}$, x intercepts are at** $\left(5 + 2 \sqrt{6} , 0\right)$

and$\left(5 - 2 \sqrt{6} , 0\right)$

graph{1/8(x-5)^2-3 [-45, 45, -22.5, 22.5]} [Ans]