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General Form $f\left(x\right)=ax^2+bx+c$

We learned that:
Quadratic Formula
If $ax^2+bx+c=0$ where $a \ne 0$, then the solutions are:

$x=$

We will derive this by completing the square.
General Form:
$\displaystyle f(x)=ax^2+bx+c$

Factor out $a$:
$\displaystyle f(x)=a \Bigl(x^2+$ $x+$ $\Bigl)$

Add and subtract the term needed to complete the square:
$\displaystyle f(x)=a \Bigl(x^2+\frac{b}{a}x+$ $\displaystyle +\frac{c}{a}-$ $\Bigl)$

Complete the square:
$\displaystyle f(x)=a \biggl\{\bigl(x+$ $\displaystyle \bigr)^2+\frac{c}{a}-\frac{b^2}{4a^2}\biggr\}$

Combine into a single fraction:
$\displaystyle f(x)=a \biggl\{\left(x+\frac{b}{2a}\right)^2+$
/ $4a^2$
$\biggr\}$

Factorize a negative:
$\displaystyle f(x)=a \left\{\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}\right\}$

Distribute $a$:
$\displaystyle f(x)=a \left(x+\frac{b}{2a}\right)^2-$
$b^2-4ac$ /

The equation of the axis of symmetry is $x=$

To find the $x$-intercepts, we let $f(x)=$ and solve for $x$.

$0\;=$
$\displaystyle a\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a}$

$\;=$
$\displaystyle a\left(x+\frac{b}{2a}\right)^2$

$\;=$
$\displaystyle \left(x+\frac{b}{2a}\right)^2$

$\pm\sqrt{b^2-4ac}$ /
$\;=$
$\displaystyle x+\frac{b}{2a}$

$\pm\sqrt{b^2-4ac}$ / $2a$
$\;=$
$x$