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Precalculus for Team-Based Inquiry Learning: PREVIEW Edition — Instructor Version

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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)?
  1. The \(y\)-intercept
  2. The \(x\)-intercepts
  3. 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 \(x^2-4x-32>0\text{.}\)
  1. \(\displaystyle (-4,8)\)
  2. \(\displaystyle [ -4,8]\)
  3. \(\displaystyle (-\infty,-4)\cup(8,\infty)\)
  4. \(\displaystyle (-\infty,-4]\cup[8,\infty)\)
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.
Figure 1.7.4. A sign chart for the function \(f(x)=x^2-4x-32\text{.}\)

Activity 1.7.5.

Solve the quadratic inequality algebraically
\begin{equation*} 2x^2-28 \lt 10x\text{.} \end{equation*}
(a)
Write your solution using interval notation.
  1. \(\displaystyle \left(-\infty,-2\right)\cup\left(7,\infty\right)\)
  2. \(\displaystyle \left(-\infty,-7\right)\cup\left(2,\infty\right)\)
  3. \(\displaystyle \left(-7,2\right)\)
  4. \(\displaystyle \left(-2,7\right)\)
Answer.
D.
(b)
Draw a number line representing your solution.
Answer.

Activity 1.7.6.

Solve the inequality
\begin{equation*} -2x^2-10x-10 \ge 6x+20\text{.} \end{equation*}
(a)
Write your solution using interval notation.
  1. \(\displaystyle [-5,-3]\)
  2. \(\displaystyle (-\infty,-5]\cup[-3,\infty)\)
  3. \(\displaystyle [3,5]\)
  4. \(\displaystyle (-\infty,3]\cup[5,\infty)\)
Answer.
A.
(b)
Draw a number line representing your solution.
Answer.

Activity 1.7.7.

Consider the rational inequality
\begin{equation*} \dfrac{x+3}{x-2} \leq 0. \end{equation*}
(a)
Use a graphing utility to graph the function \(f(x)=\dfrac{x+3}{x-2}\text{.}\) Which part of the graph represents where \(\dfrac{x+3}{x-2}\leq 0\text{?}\) Write your answer in interval notation.
Answer.
The graph is below the \(x\)-axis on the interval \([-3,2)\text{.}\)
(b)
How does the interval you found in part (a) relate to the numerator and the denominator of the function \(f(x)=\dfrac{x+3}{x-2}\text{?}\)
Answer.
The places where the numerator or denominator are zero are the potential places where the sign can change.

Activity 1.7.8.

Consider the rational inequality
\begin{equation*} \dfrac{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?
  1. \(\displaystyle f(x)=\dfrac{4x+3}{x+2}\)
  2. \(\displaystyle g(x)=\dfrac{4x+3}{x+2}-x\)
  3. \(\displaystyle h(x)=x-\dfrac{4x+3}{x+2}\)
Answer.
B or C is most helpful.
(b)
Use a graphing utility to graph the function \(g(x)=\dfrac{4x+3}{x+2}-x\text{.}\) Which part of the graph represents where \(\dfrac{4x+3}{x+2}-x\gt0\text{?}\) Write your answer in interval notation.
Answer.
The graph is above the \(x\)-axis on the interval \((-\infty, -2) \cup (-1,3)\text{.}\)
(c)
Simplify \(\dfrac{4x+3}{x+2} - x\) into a single rational expression.
  1. \(\displaystyle \dfrac{4x+3}{x+2}\)
  2. \(\displaystyle \dfrac{3x+3}{x+2}\)
  3. \(\displaystyle \dfrac{x^2+6x+3}{x+2}\)
  4. \(\displaystyle \dfrac{-x^2+2x+3}{x+2}\)
Answer.
D.
(d)
How does the interval 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 are zero are the potential places where the sign can change.
(e)
For what values is the original inequality a true statement?
  1. \(x \lt -2\) and \(-1 \lt x \lt 3\)
  2. \(-2 \lt x \lt -1\) and \(x \gt 3\)
  3. \(\displaystyle -2\lt x \lt -1\)
  4. \(\displaystyle 1 \lt x \lt 3\)
Answer.
A.
(f)
How can we express the answers to part (e) for the rational inequality using interval notation?
  1. \(\displaystyle \left(1,3\right) \)
  2. \(\displaystyle \left(-2,-1\right) \cup (3,\infty)\)
  3. \(\displaystyle \left(-2,-1\right) \)
  4. \(\displaystyle \left(-\infty,-2\right) \cup (-1,3)\)
Answer.
A.

Definition 1.7.9.

The values on the \(x\)-axis where a function is equal to zero or undefined are called partition values.

Activity 1.7.10.

Solve the rational inequality
\begin{equation*} \dfrac{x+8}{x-2} \le \dfrac{x+10}{x+5}\text{.} \end{equation*}
(a)
Write the solution using interval notation.
  1. \(\displaystyle (-\infty, -12)\cup[-5,2]\)
  2. \(\displaystyle (-\infty, -12]\cup(-5,2)\)
  3. \(\displaystyle (-12,-5]\cup[2,\infty)\)
  4. \(\displaystyle [-12,-5)\cup(2,\infty)\)
Answer.
B.
(b)
Draw a number line representing your solution.
Answer.
(c)
Compare the interval notation from Activity 1.7.8 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.

Exercises 1.7.2 Exercises

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