Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

how did you get "a"??

Answer 5

yeah your vertex is right

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

oh alright i was just making it way to complicated lol

Answer 10

but thanks

Answer 11

yeah, think easy, not hard
yw