Given a quadratic function of the form f(x) = ax^2 + bx + c, you can rewrite it in what is called "standard form," which makes it easy to see where the vertex is just by looking at the equation. Because you did not provide an example, I will demonstrate it with all variables. Can you see where I complete the square?
[tex]f(x) = ax^2 + bx+c\\
\frac{f(x)}{a}=x^2+\frac{b}{a}x+\frac{c}{a}\\
\frac{f(x)}{a}=x^2+\frac{b}{a}x+(\frac{b}{2a})^2-(\frac{b}{2a})^2+\frac{c}{a}\\
\frac{f(x)}{a}=(x+\frac{b}{2a})^2-(\frac{b}{2a})^2+\frac{c}{a}\\
\frac{f(x)}{a}=(x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{c}{a}\\
f(x)=a(x+\frac{b}{2a})^2-\frac{ab^2}{4a^2}+\frac{ca}{a}\\
f(x)=a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+\frac{4ca}{4a}\\
f(x)=a(x+\frac{b}{2a})^2+ \frac{4ac-b^2}{4a}
[/tex]
In reality, it will probably not be so complicated. It looks like a lot here, but that's because it's all variables. When you use this method, you will end up with something resembling f(x) = a(x-h)^2 + k. The vertex will be (h,k). Notice that h is SUBTRACTED from x! That means if you have a function like f(x) = (x-2)^2 + 3, the vertex is (2,3), not (-2,3).
