# Inequalities with Quadratic Functions

These are my notes on inequalities with quadratic functions.

In this section, we solve inequalities that involve quadratic functions. We will

accomplish this by using their graphs. For example, to solve the inequality

\(ax^2+bx+c>0\)

we graph the function \(f(x)=ax^2+bx+c\) and from the graph, determine where it

is above the x-axis and where f(x)>0. To solve the inequality, we graph the

function and determine where the graph is below the x-axis. If the inequality is

not strict, we include the x-intercepts in the solution.**Example 1**

Solving an Inequality

Solve the inequality \(x^2-4x-12\leq0\) and graph the solution set.

Solution:

We graph the function \(f(x)=x^2-4x-12\)

The intercepts are: x=6 and x=-2

the y-intercept is -12 with the x-intercepts at 6 and -2.

The vertex is at:

\(x=-\frac{b}{2a}=-\frac{-4}{2}=2\)

Since f(2)=-16, the vertex is (2,-16).

The graph is below the x-axis for -2<x<6. Since the original inequality is not

strict, we include the x-intercepts. The solution set is \({x|-2\leqx\leq6} or

[-2,6].**Example 2**

Solving an Inequality

Solve the inequality \(2x^2<x+10\) and graph the solution set.

Solution:

We arrange the inequality so that 0 is on the right side.

\(2x^2<x+10\)

\(2x^2-x-10<0\)

This inequality is equivalent to the one that we wish to solve.

Next, we graph the function \(f(x)=2x^2-x-10\). The intercepts are:

y=-10, x=-2, x=5/2

The vertex is at \(x=-\frac{b}{2a}=-\frac{-1}{4}=\frac{1}{4}\)

Since f(1/4)=-10.125, the vertex is at \((\frac{1}{4},-10.125)\)

The graph is below the x-axis between x=-2 and x=5/2. Since the inequality is

strict, the solution set is \({x|-2<x<\frac{5}{2}}\) or \((-2,\frac{5}{2})\).