Question

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

Answer 1

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: $y=3(x-3)^2-33$