Intermediate Algebra: Functions and Graphs

We make a table of values and plot the points. The graph is shown below.

\(x\)	\(-3\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)
\(y\)	\(5\)	\(0\)	\(-3\)	\(-4\)	\(-3\)	\(0\)	\(5\)

The figure shows the graph of \(~y=\dfrac{1}{2}x^2-7.\)

We would like to find the \(x\)-values that make \(y=8\text{.}\) The horizontal line \(y=8\) intersects the graph at two points with \(x\)-coordinates approximately \(5.5\) and \(-5.5\text{.}\) These are the solutions of the equation.

Algebraically, we solve the equation as follows.

• First, we isolate the variable. We add 7 to both sides, then multiply by 2, yielding \(~x^2=30\text{.}\)

• Because \(x\) is squared in this equation, we perform the opposite operation, or take square roots, in order to solve for \(x\text{.}\)

  \begin{align*} x^2 \amp = 30 \amp \amp \blert{\text{Take square roots of both sides.}}\\ x \amp = \pm \sqrt{30} \amp \amp \blert{\text{Remember that every positive number}}\\ \amp \amp \amp \blert{\text{has two square roots.}} \end{align*}

• We use a calculator to find that \(\sqrt{30}\) is approximately 5.477, or about 5.5, as we saw on the graph.

1. We evaluate the formula for \(t=0.5\text{.}\) We substitute \(\alert{0.5}\) for \(t\) into the formula, and simplify.

   \begin{align*} h \amp= 20-16(\alert{0.5})^2 \amp \amp \blert{\text{Compute the power.}}\\ \amp = 20-16(0.25) \amp \amp \blert{\text{Multiply, then subtract.}}\\ \amp = 20-4=16 \end{align*}

   Tux is 16 feet above the ground after 0.5 second. You can also use your calculator to simplify the expression for \(h\) by entering

   \(\qquad\qquad 20\) - \(16\) × \(0.5\) x^2 ENTER

2. We would like to find the value of \(t\) when Tux's height, \(h\text{,}\) is zero. We substitute \(h=\alert{0}\) into the equation to obtain

   \begin{equation*} \alert{0} = 20-16t^2 \end{equation*}

   To solve this equation we use extraction of roots. We first isolate \(t^2\) on one side of the equation.

   \begin{align*} 16t^2 \amp =20 \amp \amp \blert{\text{Divide by 16.}}\\ t^2 \amp = \dfrac{20}{16} = 1.25 \end{align*}

   Next, we take the square root of both sides of the equation to find

   \begin{equation*} t=\pm \sqrt{1.25} \approx \pm 1.118 \end{equation*}

   Only the positive solution makes sense here, so Tux has approximately 1.12 seconds to be in position for landing.

A graph of the Tux's height after \(t\) seconds is shown below. The points corresponding to parts (a) and (b) are labeled.

The volume of a cylinder is given by the formula \(V=\pi r^2h\text{.}\) We substitute \(\alert{582}\) for \(V\) and \(\alert{10.5}\) for \(h\text{,}\) then solve for \(r\text{.}\)

The radius of the can is 4.2 centimeters.

Because the variable we want is squared, we use extraction of roots. First, we multiply both sides by 3 to clear the fraction.

Because the radius of a cone must be a positive number, we use only the positive square root: \(\quad r = \sqrt{\dfrac{3V}{\pi h}}\text{.}\)

First, we isolate the perfect square, \(~(x-2)^2\text{.}\)

This gives us two equations for \(x\text{,}\)

The solutions are \(6\) and \(-2\text{.}\) You can check that both of these solutions satisfy the original equation.

1. We use the formula \(A=P(1+r)^n\) with \(P=3000\) and \(n=2\text{.}\) Carmella's account balance will be

   \begin{equation*} A=3000(1+r)^2 \end{equation*}

2. We substitute \(\alert{3500}\) for \(A\) in the equation.

   \begin{equation*} \alert{3500}=3000(1+r)^2 \end{equation*}

   This is a quadratic equation in the variable \(r\text{,}\) which we can solve by extraction of roots. First, we isolate the perfect square.

   \begin{align*} 3500 \amp = 3000(1+r)^2 \amp \amp \blert{\text{Divide both sides by 3000.}}\\ 1.1\overline{6} \amp = (1+r)^2 \amp \amp \blert{\text{Take square roots.}}\\ \pm 1.0801 \amp \approx 1+r \amp \amp \blert{\text{Subtract 1 from both sides.}}\\ r \approx 0.0801 \amp \text{or}~~~~r \approx -2.0801 \end{align*}

   Because the interest rate must be a positive number, we discard the negative solution. Carmella needs an account with interest rate \(r \approx 0.0801\text{,}\) or over 8%, in order to have an account balance of $3500 in two years.