how would one go about putting this: f(x)=3x^2+30x+78 into this form:
f(x)=a(x-h)^2+k??

Question

how would one go about putting this: f(x)=3x^2+30x+78 into this form: 
f(x)=a(x-h)^2+k??

anonymous · Answer

find the vertex using $-\frac{b}{2a}$

anonymous · Answer

i got the vertex it is (-5,3) i just need to know how to change the forms of the equation

anonymous · Answer

ok then it has to be 
$$f(x)=3(x+5)^2+3$$ if your vertex is right

anonymous · Answer

how did you get "a"??

anonymous · Answer

yeah your vertex is right

anonymous · Answer

i didn't really "get" it
$a$ is the leading coefficient, which in this case is $3$

anonymous · Answer

ahhh so you just have to factor it out of the original equation!! thanks alot

anonymous · Answer

if you multiply 
$$a(x-h)^2+k$$ out, the first term will be $ax^2$ right? so once you have the vertex it is easy, lots easier than completing the square

anonymous · Answer

oh alright i was just making it way to complicated lol

anonymous · Answer

but thanks

anonymous · Answer

yeah, think easy, not hard 
yw