What is the vertex of $y = (1/4)(4x – 16)^2 - 4$ ?

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Answer 1 · 2017-04-15T07:48:10

Vertex is $(4,-4)$

Vertex form of a parabola is $y = a(x+b)^2+c$

Notice that the coefficient of $x$ is 1.

In the question asked, the coefficient of $x$ is $4$ .

$y = 1/4color(red)((4x-16)^2)-4$

Simplify first: $y= 1/4color(red)((16x^2-128x+256))-4$

Factor out 16: $" "$ (the same as $4^2$ )

$y= 1/4*16color(blue)((x^2-8x+16))-4" "larr$ change to factor form

$y = 4color(blue)((x-4)^2)-4$

(we could have done this in one step at the beginning as long as the factor $4^2$ was taken out and not just $4$ )

$y = 4(x-4)^2-4$ is in vertex form.
The vertex is at $(-b,c)$

Vertex is $(4,-4)$

What is the vertex of y = (1/4)(4x – 16)^2 - 4 y = (1/4)(4x – 16)^2 - 4?