Question

Can somebody please tell me how to convert f(x) into the general, vertex form of the equation??

Answer 1

if its a quadratic, then completing the square would be useful

Answer 2

\[f(x)=ax^2+bx+c\]
\[f(x)=a(x^2+\frac bax)+c\]
\[f(x)=a(x^2+\frac bax~\pm\frac{b^2}{2a^2})+c\]
\[f(x)=a(x+\frac ba)^2~-\frac{b^2a}{2a^2}+c\]

Answer 3

4a^2, not 2a^2 forgot to square the half of it :/

Answer 4

\[f(x)=a(x^2+\frac bax~\pm\frac{b^2}{4a^2})+c\]
\[f(x)=a(x+\frac b{2a})^2~-\frac{b^2a}{4a^2}+c\]
thats better ....

Answer 5

Can you tell me in words? Haha. I don't understand all of that... @amistre64

Answer 6

i did: "if its a quadratic, then completing the square would be useful"

completing the square is a method that should be "worded out" better in your material :)

Answer 7

That's what the assignment is on is completing the square haha I just don't understand how to do that. It doesn't explain it very well where I can understand it. @amistre64

Answer 8

do you have an example of f(x) to play with?

Answer 9

its a hard method to word out, but its pretty simple to demonstrate. With a little practice it loses all of its mystery

Answer 10

if you can already find the vertex (a,b) then its just a matter of formatting the "slope"
\[y = m(x-a)^2+b\]
if you can determine a point in the curve, say (p,q), then its just a matter of working the setup
\[q = m(p-a)^2+b\]and solving for m

Answer 11

I have this:
f(x) = (x - 1)(x + 1) = x^2 - 1.

Answer 12

thats already in vertex form :)

Answer 13

Cool! :)
So... how did it get to be in the vertex form?

Answer 14

you have a squared "term" and a constant ....

if we take the vertex general format as:\[y=a(x-h)^2+k\]
and let a=1, h=0, and k=-1
\[y=1(x-0)^2+(-1)\]
\[y=x^2-1\]

Answer 15

just as a sheer coincidence, the vertex form and standard form of that setup are the same

Answer 16

Thanks :)