Question about quadratic function vertex form

Question

I'm taking Precalculus and I'm learning to graph quadratic functions by rewriting the function $f(x)=ax^2+bx+c$ in the form $f(x)=a(x-h)^2+k$. So, I tried to derive a general formula keeping the coefficients as $a$, $b$ & $c$. After doing all the algebra, I typed it into Wolfram Alpha to check, and the app said it was "not always equal" to $f(x)=ax^2+bx+c$. The formula I derived is: $$f(x)=a(x+ \frac{2b}{a})^2+(c-\frac{4b^2}{a})$$ I haven't had the opportunity to ask my professor yet, so I figured I'd ask you guys. My question is this: why is this "not always equal" as Wolfram Alpha says? Any help is appreciated.

Answer 1

You have made a mistake in your derivation.

\begin{align*} f(x)&=a\left(x^2+\frac{bx}{a}\right)+c\\ &=a\left[x^2+2(x)\left(\frac{b}{2a}\right)+\left(\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}\right]+c\\ &=a\left(x+\frac{b}{2a}\right)^2+\left(c-\frac{b^2}{4a}\right) \end{align*}