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

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Answer 1 · 2018-04-07T11:53:19

Vertex is at $(5,-3)$ y intercept is $y=1/8$ and
x intercepts are at ** $(5+ 2sqrt6 ,0) and(5- 2sqrt6 ,0)$

$y=1/8(x-5)^2-3$ Comparing with vertex form of

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

here $h=5 , k=-3 :.$ Vertex is at $(5,-3)$ , y intercept

is found by putting $x=0$ in the equation $y = 1/8(0-5)^2-3$ or

$y=25/8-3 or y= 1/8 :.$ y intercept is $y=1/8$ or at

$(0,1/8)$ , x intercept is found by putting $y=0$

in the equation $:.0 = 1/8(x-5)^2-3$ or

$1/8(x-5)^2=3 or (x-5)^2 = 24 or (x-5)= +-sqrt24$ or

$x= 5+- 2sqrt6$ , x intercepts are at** $(5+ 2sqrt6 ,0)$

and $(5- 2sqrt6 ,0)$

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

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