Intermediate Algebra: Functions and Graphs

We can get an approximate answer to this question by looking at the graph of the rocket's height, shown below.

We would like to know when the rocket's height is 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

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.

We first write the inequality with zero on one side:

We would like to find points on the graph of \(~y = x^2 - 2x - 15\) that have \(y\)-coordinates greater 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 blue on the graph lie above the \(x\)-axis and have \(y \ge 0\text{,}\) so the \(x\)-coordinates of these points are the solutions of the inequality.

Note that the solutions lie in two intervals, less than \(-3\) or greater than \(5\text{.}\) Thus, the solution is \(x \le -3\) or \(x \ge 5\text{.}\)

1. \([3,6)\text{.}\) This is called a half-open or half-closed interval.

   

2. \([-9, \infty)\text{.}\) We always use round brackets next to the symbol \(\infty\) because \(\infty\) is not a specific number and is not included in the set.

   

3. \((-\infty,1] \cup (4, \infty)\text{.}\) The word "or" describes the union of two sets.

   

4. \((-8,-5] \cup [-1,3)\text{.}\)

   

1. 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*}

2. 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 \(x = 8\text{.}\)

3. 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.

   

4. 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 \(~(-\infty,-2] \cup [8,\infty)\text{.}\)

1. 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*}

2. Consider the equation

   \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 \(c = \alert{-27,200}\text{.}\) We substitute these values into the quadratic formula to obtain

   \begin{align*} 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{align*}

   To two decimal places, the solutions to the equation are 122.54 and 277.46.

3. The graph of the equation is a parabola that opens downward, because the coefficient of \(x^2\) is negative.

   

4. 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.

   Thus, TrailGear can produce as few as 123 parkas or as many as 277 parkas per month to keep its profit above $2000.