x^2+12x+36 how do i… - QuestionCove
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