A polynomial function where the highest exponent is 2 is called a quadratic function. The graph of a quadratic function is called a parabola. It looks either like a happy smile, or a sad frown. You can see different parabolas below.

Theory

The quadratic function looks like this:

 $f\left(x\right)=a{x}^{2}+bx+c,$

where $a$, $b$ and $c$ are constants. The constants $a$ and $b$ are called coefficients. The term $a{x}^{2}$ is called a second-degree term or quadratic term, the term $bx$ is called a first-degree term or linear term, and $c$ is called the constant term.

Below you see a picture of the graphs of four different quadratic functions. Note that all have either a maximum or minimum.

Rule

### Parabola(MaximumandMinimum)

• When : The graph is smiling, and the function has a minimum (vertex).

• When : The graph is sad, and the function has a maximum (vertex).

Example 1

 $f\left(x\right)=\text{}0.5\text{}{x}^{2}˘x˘3$

Determine if the parabola has a maximum or a minimum.

You see that the coefficient in front of ${x}^{2}$ is a positive number ($a>0$). Therefore, the graph smiles and have a minimum.

Example 2

 $f\left(x\right)=-{x}^{2}˘x+12$

Determine if the parabola has a maximum or a minimum.

In this case, the coefficient in front of ${x}^{2}$ is a negative number ($a<0$). Therefore, the graph will face downwards and have a maximum, as in the figure below:

There are several methods to use to find the maximum or minimum. You can use the derivative, a sign chart, or the method that follows here:

Rule

### FindingtheMaximumorMinimumofaParabola

When you have the quadratic function

 $f\left(x\right)=a{x}^{2}+bx+c,$

you can find the $x$-values and $y$-values of the vertex like this:

The $x$-value of the maximum or minimum point:

 $x=-\frac{b}{2a}$

$y$-value of the maximum or minimum point:

 $y=f\phantom{\rule{-0.17em}{0ex}}\left(-\frac{b}{2a}\right)$

Example 3

Determine if the graph of $f\left(x\right)={x}^{2}-4x+4$ has a maximum or a minimum, and find this point

The function $f\left(x\right)$ has $a=1$, $b=-4$ and $c=4$. Since $a=1>0$ in the expression, you know that the graph is smiling and you have a minimum.

First, find the $x$-value, and then the $y$-value, of this minimum:

$\begin{array}{llll}\hfill x& =\frac{-\left(-4\right)}{2\cdot 1}=\frac{4}{2}=2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =f\left(2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={2}^{2}-\left(4\cdot 2\right)+4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =4-8+4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill x& =\frac{-\left(-4\right)}{2\cdot 1}=\frac{4}{2}=2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =f\left(2\right)={2}^{2}-4\cdot 2+4=4-8+4=0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The minimum is then $\left(2,0\right)$.

Upon inspection of $f\left(x\right)={x}^{2}-4x+4$, you can see that this is a square and therefore has only one zero. In this case, the zero and the minimum are the same point.