3.3b Sketching the Graph of Quadratic Functions

Example
Sketch the curve of the quadratic function $f(x)=x^2-<!--\; -\; -->x-<!--\; -\; -->12$

Answer:
The shape of the graph
Since the coefficient of x² is positive, hence the graph is a U shape parabola with a minimum point.

The minimum point of the graph
By completing the square
$\begin{array}{l}f(x)=x^2-x-12\\ f(x)=x^2-x+{\left(\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2-12\\ f(x)={\left(x-\frac{1}{2}\right)}^2-\frac{1}{4}-12\\ f(x)={\left(x-\frac{1}{2}\right)}^2-12\frac{1}{4}\\ \\ \text{Minimum point = }(\frac{1}{2},-12\frac{1}{4})\end{array}$

For y-intercept, x = 0
$f(0)=(0{)}^2-<!--\; -\; -->(0)-<!--\; -\; -->12=-<!--\; -\; -->12$

For x-intercept, f(x) = 0
$\begin{array}{l}f(x)=x^2-x-12\\ 0=x^2-x-12\\ (x+5)(x-6)=0\\ x=-5\text{ or }x=6\end{array}$