Inequalities with Quadratic Functions

These are my notes on inequalities with quadratic functions.

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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 \((-2,\frac{5}{2})\).