What is the vertex of y = (1/8)(x – 5)^2 - 3?

1 Answer
Yonas Yohannes
Jan 29, 2016

The vertex is v(1/20, 201/3200)

Explanation:

f(x) = y=1/8(x–5)^2−3
f(x) = 1/8 (x^2 - 10x + 1)
Now for any f(x) = y = ax^2 + bx + c " " form of the equation
The vertex, v(h, k)
h = -b/(2a); and k = f(h)
h = 1/20; k = f(1/20) = 201/3200 ~~ .063
The vertex is v(1/20, 201/3200)

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