Describing Quadratic Functions

We are given the quadratic function $y=x^2+x-12$ and we need to write it in factored form. $$\begin{array}{c}y=a(x-s)(x-t)\end{array}$$ In the equation above, $s$ and $t$ represent the solutions to the equation $y=0.$ Hence, we begin by finding the solutions using the quadratic formula. $$\begin{array}{c}x=\frac{\text{-}b\pm \sqrt{b^2-4ac}}{2a}\end{array}$$ In our case, $a=1,b=1$ and $c=\text{-}12.$ Let's substitute these values into the formula above. The $x$-intercepts of the function are $x=\frac{\text{-}1\pm 7}{2}.$ Let's separate them using the positive and negative signs.

$x=\frac{\text{-}1\pm 7}{2}$
$x=\frac{\text{-}1+7}{2}$	$x=\frac{\text{-}1-7}{2}$
$x=\frac{6}{2}$	$x=\frac{\text{-}8}{2}$
$x=3$	$x=\text{-}4$

The parabola intercepts the $x$-axis at $\text{-}4$ and $3.$ In other words, $p=\text{-}4$ and $q=3.$ We are ready to rewrite the given function in factored form. $$\begin{array}{c}f(x)=(x-(\text{-}4))(x-3)=(x+4)(x-3)\end{array}$$ Next, we find the axis of symmetry by finding the midpoint between the $x$-intercepts. $$\begin{array}{c}x=\frac{\text{-}4+3}{2}=\text{-}\frac{1}{2}\end{array}$$ From the above, we also know that the $x$-coordinate of the vertex is $\text{-}\frac{1}{2}$ and the $y$-coordinate is equal to $f\left(\text{-}\frac{1}{2}\right).$ Let's find the $y$-coordinate. We now know that the vertex is at $(\text{-}\frac{1}{2},\text{-}\frac{49}{4}).$ Finally, we can find the $y$-intercept by setting $x=0$ in $y=x^2+x-12.$ $$\begin{array}{c}y=0^2+0-12=\text{-}12\end{array}$$ The graph intercepts the $y$-axis at $(0,\text{-}12).$ With all this information, we can graph the function.