How do you find the vertex for $2x^-16x+4$ ?

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Answer 1 · 2015-06-27T21:50:23

I think you mean $2x^2-16x+4$

$2x^2-16x+4 = 2(x-4)^2-28$

This is in vertex form allowing us to read the vertex as $(4, -28)$

Let $f(x) = 2x^2-16x+4$

Then $f(x) = 2(x^2-8x+16)-32+4 = 2(x-4)^2-28$

$f(x) = 2(x-4)^2-28$ is in vertex form, being of the form:

$f(x) = a(x-x_1)+y_1$ ,

where $(x_1, y_1) = (4, -28)$ is the vertex and $a$ is some constant.

More explicitly:

The minimum value of $f(x)$ occurs when $(x-4) = 0$ , that is when $x = 4$ .

$f(4) = 2(4-4)^2-28 = 0-28 = -28$

So the vertex is at $(4, -28)$

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