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

y-intercept is $\frac{1}{8} \mathmr{and} .125$ ; x -intercepts are $5 + 2 \sqrt{6} \mathmr{and} 9.9 \mathmr{and} 5 - 2 \sqrt{6} \mathmr{and} .1$
Here Vertex is $\left(5 , - 3\right)$ y-intercept: putting x=0 ; $y = \frac{1}{8} \cdot 25 - 3 = \frac{1}{8}$
To find x intercepts putting y=0 we get $\frac{1}{8} \cdot {\left(x - 5\right)}^{2} = 3 \mathmr{and} {\left(x - 5\right)}^{2} = 24$
$\therefore \left(x - 5\right) = 2 \sqrt{6} \mathmr{and} \left(x - 5\right) = - 2 \sqrt{6} \mathmr{and} x = 5 + 2 \sqrt{6} \mathmr{and} x = 5 - 2 \sqrt{6}$ graph{1/8*(x-5)^2-3 [-20, 20, -10, 10]} [answer]