Formula for Xo and Yo for graph of quadratic function

[Callmelucky]

Homework Statement

Shouldn't it be $\frac{b^{2}}{4a^{2}}$ after raising (b/2a) to power of 2?

Relevant Equations

f(x)= ax^2 + bx + c, f(x)=a(X-Xo)^2 + Yo

Can someone please tell me where I am wrong. I am learning how to write $a^{2} + bx + c$ in this form $f(x)= a(X-X_0)^{2} +Y_0$.
The method used in my textbook is a reduction to the perfect square. And it goes like this:

$f(x)=ax^2+bx+c$
$=a[x^2+\frac{b}{a}x]+c$
$=a\left [ x^2+\frac{b}{a}x+\left ( \frac{b}{2a} \right )^2-\left ( \frac{b}{2a} \right )^2 \right ]+c$
$=a\left ( x+\frac{b}{2a} \right )^2 -\frac{b^2}{4a}+c$
$f(x)=a\left ( x+\frac{b}{2a} \right )^2 + \frac{4ac-b^2}{4a}$

Now my question is shouldn't in the last two steps after squaring $\frac{b}{2a}$ be $\frac{b^2}{4a^2}$ and not $\frac{b^2}{4a}$ as it is written in my textbook, I have also checked on the internet but the answer is the same, although I couldn't find why.

Thank you.

 

[DaveE]

You have forgotten to account for the factor of $a$ on the far left. So, ignoring the other terms you get $a[... -(\frac{b^2}{4a^2})]... = ... -(\frac{b^2}{4a}) ...$

 

[Callmelucky]

You have forgotten to account for the factor of $a$ on the far left. So, ignoring the other terms you get $a[... -(\frac{b^2}{4a^2})]... = ... -(\frac{b^2}{4a}) ...$

oh, yeah. Thank you.