Objectives
- Solve quadratic inequalities and express the solution graphically and with interval notation. Solve rational inequalities and express the solution graphically and using interval notation.
Section 1.7 Quadratic and Rational Inequalities (EQ7)
Subsection 1.7.1 Activities
Remark 1.7.1.
In Section 1.5 and Section 1.6 we learned how to solve quadratic and rational equations. In this section, we use these skills to solve quadratic and rational inequalities.
Activity 1.7.2.
Consider the quadratic inequality
\begin{equation*}
x^2-4x-32\gt0\text{.}
\end{equation*}
(a)
Use a graphing utility to graph the function \(f(x)=x^2-4x-32\text{.}\) Which part of the graph represents where \(x^2-4x-32\gt0\text{?}\)
Answer.
(b)
Which pieces of information about \(f(x)=x^2-4x-32\) were needed to answer part (a)?
- The \(y\)-intercept
- The \(x\)-intercepts
- The minimum value
Answer.
B.
(c)
Use algebra to find the \(x\)-intercepts of \(f(x)=x^2-4x-32\) and mark them on a number line. Then, shade the part of the number line where \(x^2-4x-32>0\text{.}\)
Answer.
(d)
Now use interval notation to express where \(x^2-4x-32>0\text{.}\)
- \(\displaystyle (-4,8)\)
- \(\displaystyle [ -4,8]\)
- \(\displaystyle (-\infty,-4)\cup(8,\infty)\)
- \(\displaystyle (-\infty,-4]\cup[8,\infty)\)
Answer.
C.
Definition 1.7.3.
A sign chart is a number line representing the \(x\)-axis that shows where a function is positive or negative. Instead of shading, which can be ambiguous, it is often decorated with a ’+’ or a ’-’ to indicate which regions are positive or negative. For example, a sign chart for \(f(x)=x^2-4x-32\) is below.
Activity 1.7.5.
Solve the quadratic inequality algebraically
\begin{equation*}
2x^2-28 \lt 10x\text{.}
\end{equation*}
Write your solution using interval notation.
- \(\displaystyle \left(-\infty,-2\right)\cup\left(7,\infty\right)\)
- \(\displaystyle \left(-\infty,-7\right)\cup\left(2,\infty\right)\)
- \(\displaystyle \left(-7,2\right)\)
- \(\displaystyle \left(-2,7\right)\)
Answer.
D.
Activity 1.7.6.
Solve the inequality
\begin{equation*}
-2x^2-10x-10 \ge 6x+20\text{.}
\end{equation*}
Write your solution using interval notation.
- \(\displaystyle [-5,-3]\)
- \(\displaystyle (-\infty,-5]\cup[-3,\infty)\)
- \(\displaystyle [3,5]\)
- \(\displaystyle (-\infty,3]\cup[5,\infty)\)
Answer.
A.
Activity 1.7.7.
Consider the rational inequality
\begin{equation*}
\frac{4x+3}{x+2} \gt x\text{.}
\end{equation*}
(a)
For which of the following functions will a graph help us solve the rational inequality above?
- \(\displaystyle f(x)=\frac{4x+3}{x+2}\)
- \(\displaystyle g(x)=\frac{4x+3}{x+2}-x\)
- \(\displaystyle h(x)=x-\frac{4x+3}{x+2}\)
Answer.
B or C is most helpful.
(b)
Use a graphing utility to graph the function \(g(x)=\frac{4x+3}{x+2}-x\text{.}\) Which part of the graph represents where \(\frac{4x+3}{x+2}-x\gt0\text{?}\)
(c)
Simplify \(\frac{4x+3}{x+2} - x\) into a single rational expression.
- \(\displaystyle \frac{4x+3}{x+2}\)
- \(\displaystyle \frac{3x+3}{x+2}\)
- \(\displaystyle \frac{x^2+6x+3}{x+2}\)
- \(\displaystyle \frac{-x^2+2x+3}{x+2}\)
Answer.
D.
(d)
How do these values you found in part (b) relate to the numerator and the denominator of the combined rational function in part (c)?
Hint.
Factor the numerator.
Answer.
The places where the numerator or denominator vanishes are the potential places where the sign can change.
(e)
For what values is the original inequality a true statement?
- \(x \lt -2\) and \(-1 \lt x \lt 3\)
- \(-2 \lt x \lt -1\) and \(x \gt 3\)
- \(\displaystyle -2\lt x \lt -1\)
- \(\displaystyle 1 \lt x \lt 3\)
Answer.
A.
(f)
How can we express the answers to part (e) for the rational inequality using interval notation?
- \(\displaystyle \left(-\infty,-2\right) \cup (-1,3)\)
- \(\displaystyle \left(-2,-1\right) \cup (3,\infty)\)
- \(\displaystyle \left(-2,-1\right) \)
- \(\displaystyle \left(1,3\right) \)
Answer.
A.
Definition 1.7.8.
The values on the \(x\)-axis where a function is equal to zero or undefined are called partition values.
Activity 1.7.9.
Solve the rational inequality
\begin{equation*}
\frac{x+8}{x-2} \le \frac{x+10}{x+5}\text{.}
\end{equation*}
(a)
Write the solution using interval notation.
- \(\displaystyle (-\infty, -12)\cup[-5,2]\)
- \(\displaystyle (-\infty, -12]\cup(-5,2)\)
- \(\displaystyle (-12,-5]\cup[2,\infty)\)
- \(\displaystyle [-12,-5)\cup(2,\infty)\)
Answer.
B.
(b)
Compare the interval notation from Activity 1.7.7 to the interval notation for this activity. When do we include the partition values in the answer with a bracket?
Answer.
Values that make the function undefined cannot be solutions, while values that make the simplified rational function zero are solutions.
Subsection 1.7.2 Videos
It would be great to include videos down here, like in the Calculus book!
