What is the vertex form of $y= (x + 2)(x - 6)$ ?

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Answer 1 · 2016-05-14T19:17:44

$y = (x-2)^2-16$

or if you want it in exactly $y = a(x-h)^2+k$ form:

$y = 1(x-2)^2+(-16)$

$y = (x+2)(x-6) = (x-2)^2-16$

How did I get that?

Since we have factors $(x+2)$ and $(x-6)$ this parabola will intercept the $x$ axis at $(-2, 0)$ and $(6, 0)$ .

Then the vertical axis of symmetry must lie half way between these, making it the line $x = 2$ , which is also the $x$ coordinate of the vertex.

So we must have:

$y = (x-2)^2+c$

for some constant $c$ .

Then putting $x=2$ in the original equation, we find:

$(x+2)(x-6) = (2+2)(2-6) = 4*(-4) = -16$

So $c = -16$

The vertex of the parabola is at $(2, -16)$

What is the vertex form of y= (x + 2)(x - 6)y= (x + 2)(x - 6)?