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)#