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Write 3x^2-18x-6 in vertex form

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The standard form of a quadratic equation is
\displaystyle{ y=ax^2+bx+c, while the vertex form is:


y=a(x-h)^2+k, where (h, k) is the vertex of the parabola.

What we want is to write
\displaystyle{ y=3x^2-18x-6 as
y=a(x-h)^2+k

First, we note that all the three terms have a factor of 3, so we factorize it and write:


\displaystyle{ y=3(x^2-6x-2).


Second, we notice that
x^2-6x are the terms produced by
(x-3)^2=x^2-6x+9, without the 9. So we can write:


x^2-6x=(x-3)^2-9, and substituting in
\displaystyle{ y=3(x^2-6x-2) we have:


\displaystyle{ y=3(x^2-6x-2)=3[(x-3)^2-9-2]=3[(x-3)^2-11].

Finally, distributing 3 over the two terms in the brackets we have:


y=3[x-3]^2-33.


Answer:
y=3(x-3)^2-33
User Tchan
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