Question

x^2+12x+36 how do i put this in vertex form? can someone show me the steps?

Answer 1

\[x^2+12x+36\\=(x^2+12x +(\frac{12}{2})^2-(\frac{12}{2})^2)+36\\=(x^2+12x+(\frac{12}{2})^2)+36-(\frac{12}{2})\\=(x+6)^2+30\]

Answer 2

oops

Answer 3

first coordinate of the vertex is \(-\frac{b}{2a}\) which in your case is \(-\frac{12}{2}=-6\) so you know it looks like
\[(x+6)^2+k\] to find \(k\) replace \(x\) by \(-6\) to see what you get

Answer 4

\[(x^2+12x+(\frac{12}{2})^2)+36-(\frac{12}{2})^2\\=(x+6)^2\]

Answer 5

would the vertex be considered an intercept?

Answer 6

not in general
but in this case it is, because you have a perfect square, namely \((x+6)^2\) so it is only zero if \(x=-6\)

Answer 7

the vertex isn't an intercept unless the vertex is on the x axis or y axis

Answer 8

@satellite73 could you take a look at something for me and tell me where my mistake is? I need a fresh pair of eyes.

Answer 9

thank you guys! and the domain is all real numbers, correct?

Answer 10

yup! there are no numbers that cause any problems :)