Question

I am working on Analyzing Graphs of Quadratic Functions, the problem I am stuck on is f(x)=x^2/2+4x+4. I think the area I am messing up on is the squaring. Someone please putting this in the most simplest form. Algebra is my worst enemy.

Answer 1

what exactly are you supposed to do?

Answer 2

put it in the form f(x)=a(x-h)^2+k. find the vertex and graph it.

Answer 3

is it
\[y=\frac{x^2}{2}+4x+4\]?

Answer 4

that's the original function. but I don't know how to work it out to because an order pair

Answer 5

if you want to write in vertex form first factor out the \(\frac{1}{2}\) from the first two terms and write
\[y=\frac{1}{2}(x^2+2x)+4\]

Answer 6

easiest way to find the vertex is to use \(-\frac{b}{2a}\) for the first coordinate

Answer 7

in your example it is
\[-\frac{4}{2\times \frac{1}{2}}=-4\]

Answer 8

i wrote something stupid \[y=\frac{1}{2}(x^2+2x)+4\] if you factor out the \(\frac{1}{2}\) you actually get \[y=\frac{1}{2}(x^2+\color{red}8x)+4\]

Answer 9

thanks I got it now.