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

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Answer 1 · 2016-06-10T19:16:46

the vertex is at $(5,-3)$

Your quadratic equation is of the form $y=a (x-h)^2+k$
The vertex is at the point $(h,k)$

$y= 1/8 (x-5)^2 + (-3)$

the vertex is at $(5,-3)$
graph{1/8(x-5)^2-3 [-2.03, 14.67, -6.53, 1.814]}

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