Subsection Solving Inequalities Graphically
In
Chapter 1, we used graphs to solve equations and inequalities. The graphing technique is especially helpful for solving quadratic inequalities.
Example 6.89.
The Chamber of Commerce in River City plans to put on a Fourth of July fireworks display. City regulations require that fireworks at public gatherings explode higher than \(800\) feet above the ground. The mayor particularly wants to include the Freedom Starburst model, which is launched from the ground. Its height after t seconds is given by
\begin{equation*}
h = f(t) = 256t - 16t^2
\end{equation*}
When should the Starburst explode in order to satisfy the safety regulation?
Solution.
We can get an approximate answer to this question by looking at the graph of the rocket’s height, shown below.
When is the rocket’s height greater than 800 feet, or, in mathematical terms, for what values of \(t\) is \(h \gt 800\text{?}\) The answer to this question is the solution of the inequality
\begin{equation*}
256t - 16t^2 \gt 800
\end{equation*}
Points on the graph with \(h \gt 800\) are shown in color, and the \(t\)-coordinates of those points are marked on the horizontal axis. If the Freedom Starburst explodes at any of these times, it will satisfy the safety regulation.
From the graph, the safe time interval runs from approximately \(4.25\) seconds to \(11.75\) seconds after launch. The solution of the inequality is the set of all \(t\)-values greater than \(4.25\) but less than \(11.75\text{.}\)
Checkpoint 6.91. QuickCheck 1.
What are the solutions of the inequality \(x^2+1 \lt 0\) ?
\(\displaystyle (-1,1)\)
\(\displaystyle (-\infty,-1) \cup (1,\infty)\)
All real numbers
No solution
Technology 6.92. Solving an Inequality With a Graphing Calculator.
You can use your graphing calculator to solve the problem in
Example 6.89. Graph the two functions
\begin{equation*}
\begin{aligned}[t]
Y_1 \amp= 256X - 16X^2\\
Y_2 \amp= 800
\end{aligned}
\end{equation*}
on the same screen. Use
WINDOW settings to match the graph in
Example 6.89.
Then use the intersect feature to find the \(x\)-coordinates of the points where the two graphs intersect (there are two of them). These points will have \(y\)-coordinates of \(800\text{.}\) The parabola is above the line, so \(h \gt 800\) when \(t\) is between these two \(x\)-values. To two decimal places, you can see that \(h \gt 800\) when \(4.26 \lt t \lt 11.74\text{.}\)
Checkpoint 6.93. Practice 1.
Graph the function \(y = x^2 - 2x - 9\) in the window
\begin{equation*}
\begin{aligned}[t]
\text{Xmin} \amp = -0.4 \amp\amp \text{Xmax} = 9.4\\
\text{Ymin} \amp = -10 \amp\amp \text{Ymax} = 10
\end{aligned}
\end{equation*}
Use the graph to solve the inequality \(x^2 - 2x - 9 \ge 6\text{.}\)
Solution.
\(x\le -3\) or \(x\ge 5\)
In
Example 6.89, we solved the inequality
\(256t - 16t^2 \gt 800\) by comparing points on the graph of
\(h = 256 - 16t^2\) with points on the line
\(h = 800\text{.}\) If one side of an inequality is zero, we can compare points on the graph with the line
\(y = 0\text{,}\) which is the
\(x\)-axis.
Example 6.94.
Consider the graph of \(y = x^2 - 4\text{.}\) Find the solutions of the following equations and inequalities.
\(\displaystyle x^2 - 4 = 0\)
\(\displaystyle x^2 - 4 \lt 0\)
\(\displaystyle x^2 - 4 \gt 0\)
Solution.
Look at the graph of \(y = x^2 - 4\) shown below. When we substitute a value of \(x\) into the expression \(x^2 - 4\text{,}\) the result is either positive, negative, or zero.
(You can see this more clearly if you compute a few values yourself to complete the table below. Your table should agree with the coordinates of points on the graph.)
| \(x\) |
\(-3\) |
\(-2\) |
\(-1\) |
\(0\) |
\(1\) |
\(2\) |
\(3\) |
| \(y\) |
\(\hphantom{bla}\) |
\(\hphantom{bla}\) |
\(\hphantom{bla}\) |
\(\hphantom{bla}\) |
\(\hphantom{bla}\) |
\(\hphantom{bla}\) |
\(\hphantom{bla}\) |
We will use the graph to solve the given equation and inequalities.
First, locate the two points on the graph where \(y = 0\text{.}\) These points are \((-2, 0)\) and \((2, 0)\text{.}\) Their \(x\)-coordinates, \(-2\) and \(2\text{,}\) are the solutions of the equation \(x^2 - 4 = 0\text{.}\) The two points divide the \(x\)-axis into three sections, which are labeled on the graph. On each of these sections, the value of \(x^2 - 4\) is either always positive or always negative.
To solve \(x^2 - 4 \lt 0\text{,}\) find the points on the graph where \(y \lt 0\text{,}\) that is, below the \(x\)-axis. These points have \(x\)-coordinates between \(-2\) and \(2\) (labeled section II on the figure). Thus, the solution to the inequality \(x^2 - 4 \lt 0\) is \(-2 \lt x \lt 2\text{.}\)
To solve \(x^2 - 4 \gt 0\text{,}\) locate the points on the graph where \(y \gt 0\text{,}\) or above the \(x\)-axis. Points with positive \(y\)-values correspond to two sections of the \(x\)-axis, labeled I and III on the figure. In section I, \(x \lt -2\text{,}\) and in section III, \(x \gt 2\text{.}\) Thus, the solution to the inequality \(x^2 - 4 \gt 0\) includes all values of \(x\) for which either \(x \lt -2\) or \(x \gt 2\text{.}\)
Checkpoint 6.96. Practice 2.
Use a graph of \(y = 6 - x^2\) to solve the inequalities.
\(\displaystyle 6 - x^2 \gt 0\)
\(\displaystyle 6 - x^2 \le 0\)
Solution.
\(\displaystyle -\sqrt{6}\lt x \lt \sqrt{6} \)
\(x \le -\sqrt{6} \) or \(x\ge \sqrt{6} \)
Checkpoint 6.97. QuickCheck 2.
Which points on the parabola \(~y=ax^2+bx+c~\) do we need to know to solve the inequality \(~ax^2+bx+c \gt 0\) ?
The \(x\)-intercepts
The \(y\)-intercept
The vertex
All of these
Because it is relatively easy to decide whether the \(y\)-coordinate of a point on a graph is positive or negative (the point lies above the \(x\)-axis or below the x-axis), we often rewrite a given inequality so that one side is zero.
Example 6.98.
Use a graph to solve \(~x^2 - 2x - 3 \le 12\)
Solution.
We first write the inequality with zero on one side:
\begin{equation*}
x^2 - 2x - 15 \le 0\text{.}
\end{equation*}
We would like to find points on the graph of \(y = x^2 - 2x - 15\) that have \(y\)-coordinates less than or equal to zero. A graph of the equation is shown below.
You can check that the \(x\)-intercepts of the graph are \(-3\) and \(5\text{.}\) The points shown in red on the graph lie below the \(x\)-axis and have \(y \le 0\text{,}\) so the \(x\)-coordinates of these points are the solutions of the inequality. All of these points have \(x\)-coordinates between \(-3\) and \(5\text{.}\) Thus, the solution is \(-3\le x\le 5\text{,}\) or in interval notation, \([-3, 5]\text{.}\)
Checkpoint 6.99. Practice 3.
Follow the steps below to solve the inequality \(~36 + 6x - x^2\le 20\text{.}\)
Rewrite the inequality so that the right side is zero.
Graph the equation \(y = 16 + 6x - x^2\text{.}\)
Locate the points on the graph with \(y\)-coordinate less than zero, and mark the \(x\)-coordinates of the points on the \(x\)-axis.
Write the solution with interval notation.
Solution.
\(\displaystyle 16+6x-x^2\le 0\)
See graph
\(\displaystyle (-\infty,-2] \cup [8,\infty)\)
Checkpoint 6.100. Pause and Reflect.
Explain why you cannot write the solutions to \(x^2-4 \gt 0\) as a single inequality.
Subsection Solving Quadratic Inequalities Algebraically
Although a graph is very helpful in solving inequalities, it is not completely necessary. Every quadratic inequality can be put into one of the forms
\begin{equation*}
\begin{aligned}[t]
ax^2 + bx + c \amp\lt 0\text{,}
\amp\hphantom{blank} ax^2 + bx + c \amp\gt 0\\
ax^2 + bx + c \amp\le 0\text{,}
\amp\hphantom{blank} ax^2 + bx + c \amp\ge 0\\
\end{aligned}
\end{equation*}
All we really need to know is whether the corresponding parabola \(y = ax^2 + bx + c\) opens upward or downward. Consider the parabolas shown below.
The parabola in figure (a) opens upward. It crosses the \(x\)-axis at two points, \(x = r_1\) and \(x = r_2\text{.}\) At these points, \(y = 0\text{.}\)
The graph lies below the \(x\)-axis between \(r_1\) and \(r_2\text{,}\) so the solutions to the inequality \(y\lt 0\) lie between \(r_1\) and \(r_2\text{.}\)
The graph lies above the \(x\)-axis for \(x\)-values less than \(r_1\) or greater than \(r_2\text{,}\) so the solutions to the inequality \(y\gt 0\) are \(x\lt r_1\) or \(x\gt r_2\text{.}\)
If the parabola opens downward, as in figure (b), the situation is reversed. The solutions to the inequality \(y\gt 0\) lie between the \(x\)-intercepts, and the solutions to \(y\lt 0\) lie outside the \(x\)-intercepts.
From the graphs, we see that the \(x\)-intercepts are the boundary points between the portions of the graph with positive \(y\)-coordinates and the portions with negative \(y\)-coordinates. To solve a quadratic inequality, we need only locate the \(x\)-intercepts of the corresponding graph and then decide which intervals of the \(x\)-axis produce the correct sign for \(y\text{.}\)
To Solve a Quadratic Inequality Algebraically:.
Write the inequality in standard form: One side is \(0\text{,}\) and the other has the form \(ax^2 + bx + c\text{.}\)
Find the \(x\)-intercepts of the graph of \(y = ax^2 + bx + c\) by setting \(y = 0\) and solving for \(x\text{.}\)
Make a rough sketch of the graph, using the sign of \(a\) to determine whether the parabola opens upward or downward.
Decide which intervals on the \(x\)-axis give the correct sign for \(y\text{.}\)
Example 6.101.
Solve the inequality \(~36 + 6x - x^2\le 20~\) algebraically.
Solution.
We subtract \(20\) from both sides of the inequality so that we have \(0\) on the right side.
\begin{equation*}
16 + 6x - x^2\le 0
\end{equation*}
Consider the equation \(y = 16 + 6x - x^2\text{.}\) To locate the \(x\)-intercepts, we set \(y = 0\) and solve for \(x\text{.}\)
\begin{align*}
16 + 6x - x^2 \amp= 0 \amp\amp \blert{\text{Multiply each term by }-1.}\\
x^2 - 6x - 16 \amp = 0 \amp\amp \blert{\text{Factor the left side.}}\\
(x - 8)(x + 2)\amp= 0\amp\amp \blert{\text{Apply the zero-factor principle.}}\\
x - 8 = 0 ~~~\text{ or }~~~x + 2 \amp= 0\\
x = 8 ~~~\text{ or }~~~ x \amp= -2
\end{align*}
The \(x\)-intercepts are \(x = -2\) and \(x = 8\text{.}\)
-
Make a rough sketch of the graph of \(y = 16 + 6x - x^2\text{,}\) as shown below. Because \(a = -1 \lt 0\text{,}\) the graph is a parabola that opens downward.
We are interested in points on the graph for which \(y\le 0\text{.}\) The points with negative \(y\)-coordinates (that is, points below the \(x\)-axis) lie outside the \(x\)-intercepts of the graph, so the solution of the inequality is \(x\le -2\) or \(x\ge 8\text{.}\) Or, using interval notation, the solution is \((-\infty,-2] \cup [8,\infty)\text{.}\)
Checkpoint 6.102. QuickCheck 3.
Why do we need to know whether tha parabola in QuickCheck 2 opens up or down?
To decide whether to use \(\gt\) or \(\lt\) in the solution.
To decide whether the solutions lie between the \(x\)-intercepts or outside them.
To decide whether the solutions are positive or negative.
To help us find the \(x\)-intercepts.
Checkpoint 6.104. Practice 4.
Solve \(x^2 \lt 20\text{.}\)
Write the inequality in standard form.
Find the \(x\)-intercepts of the corresponding graph. Use extraction of roots.
Make a rough sketch of the graph.
Decide which intervals on the \(x\)-axis give the correct sign for \(y\text{.}\)
Solution.
\(-\sqrt{20} \lt x \lt \sqrt{20} \)
If we cannot find the \(x\)-intercepts of the graph by factoring or extraction of roots, we can use the quadratic formula.
Example 6.105.
TrailGear, Inc. manufactures camping equipment. The company finds that the profit from producing and selling \(x\) alpine parkas per month is given, in dollars, by
\begin{equation*}
P = -0.8x^2 + 320x - 25,200
\end{equation*}
How many parkas should the company produce and sell each month if it must keep the profits above $\(2000\text{?}\)
Solution.
We would like to solve the inequality
\begin{equation*}
-0.8x^2 + 320x - 25,200\gt 2000
\end{equation*}
or, subtracting \(2000\) from both sides,
\begin{equation*}
-0.8x^2 + 320x - 27,200 \gt 0
\end{equation*}
Consider the function
\begin{equation*}
y = -0.8x^2 + 320x - 27,200
\end{equation*}
We locate the \(x\)-intercepts of the graph by setting \(y = 0\) and solving for \(x\text{.}\) We will use the quadratic formula to solve the equation
\begin{equation*}
-0.8x^2 + 320x - 27,200 = 0
\end{equation*}
so \(a = \alert{-0.8}\text{,}\) \(b = \alert{320}\text{,}\) and \(c = \alert{-27,200}\text{.}\)
\begin{equation*}
\begin{aligned}[t]
x \amp=\frac{-(\alert{320})\pm\sqrt{(\alert{320})^2 - 4(-0.8)(\alert{-27,200})}} {2(\alert{-0.8})}\\
\amp=\frac{-320\pm\sqrt{102,400 - 87,040}}{-1.6}\\
\amp=\frac{-320\pm\sqrt{15,360}}{-1.6}
\end{aligned}
\end{equation*}
To two decimal places, the solutions to the equation are \(122.54\) and \(277.46\text{.}\)
-
The graph of the function is a parabola that opens downward, because the coefficient of \(x^2\) is negative.
The graph lies above the \(x\)-axis, and hence \(y \gt 0\text{,}\) for \(x\)-values between the two \(x\)-intercepts, that is, for \(122.54 \lt x \lt 277.46\text{.}\) Because we cannot produce a fraction of a parka, we restrict the interval to the closest whole number \(x\)-values included, namely \(123\) and \(277\text{.}\) Thus, TrailGear can produce as few as \(123\) parkas or as many as \(277\) parkas per month to keep its profit above $\(2000\text{.}\)
Checkpoint 6.106. QuickCheck 4.
What does the notation \([-3,3]\) mean?
The point with \(x\)-coordinate 3 and \(y\)-coordinate 3.
\(x = 3\) or \(x = -3\)
All real numbers between \(-3\) and \(3\text{,}\) excluding the endpoints.
All real numbers between \(-3\) and \(3\text{,}\) including the endpoints.
Checkpoint 6.107. Practice 5.
Solve the inequality \(10 - 8x + x^2\gt 4\text{.}\)
Write the inequality in standard form.
Find the \(x\)-intercepts of the corresponding graph. Use extraction of roots.
Make a rough sketch of the graph.
Decide which intervals on the \(x\)-axis give the correct sign for \(y\text{.}\)
Solution.
\((-\infty, 4 - \sqrt{10}] \cup [4 + \sqrt{10},\infty)\)
Checkpoint 6.108. Pause and Reflect.
Explain how to use a graph to solve \(~x^2+4x-77 \le 0\text{.}\)
