OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

oops

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

would the vertex be considered an intercept?

4 years ago
OpenStudy (anonymous):

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 \(x=-6\)

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago