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Section 1.3 Intercepts

Subsection 1. Graph a linear equation by the intercept method

To graph a line by the intercept method, we find the \(x\)- and \(y\)-intercepts of the line and plot those points.

Example 1.46.

Graph the equation \(3x+2y=7\) by the intercept method.

Solution

First, we find the \(x\)- and \(y\)-intercepts of the graph. To find the \(y\)-intercept, we substitute \(0\) for \(x\) and solve for \(y\text{:}\)

\begin{align*} 3(\alert{0})+2y \amp =7 \amp \amp \blert{\text{Simpify the left side.}}\\ 2y \amp =7 \amp \amp \blert{\text{Divide both sides by 2.}}\\ y \amp =\dfrac{7}{2}=3\dfrac{1}{2} \end{align*}

The \(y\)-intercept is the point \(\left(0, 3\dfrac{1}{2}\right)\text{.}\) To find the \(x\)-intercept, we substitute \(0\) for \(y\) and solve for \(x\text{:}\)

\begin{align*} 3x+2(\alert{0}) \amp =7 \amp \amp \blert{\text{Simpify the left side.}}\\ 3x \amp =7 \amp \amp \blert{\text{Divide both sides by 3.}}\\ x \amp =\dfrac{7}{3}=2\dfrac{1}{3} \end{align*}

The \(x\)-intercept is the point \(\left(2\dfrac{1}{3}, 0\right)\text{.}\)

A table with the two intercepts is shown below. We plot the intercepts and connect them with a straight line.

\(x\) \(y\)
\(0\) \(3\dfrac{1}{2}\)
\(2\dfrac{1}{3}\) \(0\)
line

Subsubsection Exercises

Graph the line \(y=\dfrac{-4}{3}x+8\) by the intercept method.

\(x\) \(y\)
\(~~~~~~~~\) \(~~~~~~~~\)
\(~~~~~~~~\) \(~~~~~~~~\)
8 by 8 grid
Answer
line

Graph the line \(\dfrac{x}{6}+\dfrac{y}{8}=-1\) by the intercept method.

\(x\) \(y\)
\(~~~~~~~~\) \(~~~~~~~~\)
\(~~~~~~~~\) \(~~~~~~~~\)
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Answer
line

Subsection 2. Interpret the intercepts

The values of the variables at the intercepts often tell us something important about a linear model

Example 1.49.

The temperature, \(T\text{,}\) in Nome was \(-12 \degree\) at noon and has been rising at a rate of \(2 \degree\) per hour all day.

  • Write and graph an equaton for \(T\) in terms of \(h\text{,}\) the number of hours after noon.
  • Find the intercepts of the graph and interpret their meaning in the context of the problem situation.
Solution

An equation for \(T\) at time \(h\) is

\begin{equation*} T=-12 +2h \end{equation*}

To find the \(T\)-intercept, we set \(h=0\) and solve for \(T\text{.}\)

\begin{equation*} T=-12+2(\alert{0})=-12 \end{equation*}
Temperature in Nome

The \(T\)-intercept is \((0,-12)\text{.}\) This point tells us that when \(h=0, T=-12\text{,}\) or the temperature at noon was \(-12 \degree\text{.}\) To find the \(h\)-intercept, we set \(T=0\) and solve for \(h\text{.}\)

\begin{align*} \alert{0} \amp = -12+2h \amp \amp \blert{\text{Add 12 to both sides.}}\\ 12 \amp = 2h \amp \amp \blert{\text{Divide both sides by 2.}}\\ 6 \amp =h \end{align*}

The \(h\)-intercept is the point \((6,0)\text{.}\) This point tells us that when \(h=6, T=0\text{,}\) or the temperature will reach zero degrees at six hours after noon, or 6 pm.

Subsubsection Exercises

Sheri bought a bottle of multivitamins for her family. The number of vitamins lt in the bottle after \(d\) days is given by

\begin{equation*} N=300-5d \end{equation*}
  1. Find the intercepts and use them to make a graph of the equation.
    \(d\) \(N\)
    \(~~~~~~~~\) \(~~~~~~~~\)
    \(~~~~~~~~\) \(~~~~~~~~\)
  2. Explain what each intercept tells us about the vitamins.
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Answer
grid
  • \((0,300)\) There were 300 vitamins to start.
  • \((60,0)\) The vitamin bottle is empty after 60 days.

Delbert bought some equipment and went into the dog-grooming business. His profit is increasing according to the equation

\begin{equation*} P=-600+40d \end{equation*}

where \(d\) is the number of dogs he has groomed.

  1. Find the intercepts and use them to make a graph of the equation.
    \(d\) \(P\)
    \(~~~~~~~~\) \(~~~~~~~~\)
    \(~~~~~~~~\) \(~~~~~~~~\)
  2. Explain what each intercept tells us about Delbert's dog-grooming business.
grid
Answer
grid
  • \((0,-600)\) To start, Delbert's profit is \(-$600\text{.}\) (He is $600 in debt.)
  • \((15,0)\) Delbert breaks even after grooming 15 dogs.

Subsection 3. Solve an equation for one of the variables

It is usually easier to study a model and draw its graph if it is in the form

\begin{equation*} y = \text{starting value} + \text{rate} \times x \end{equation*}

To put an equation into this form, we want to "isolate" the output variable on one side of the equation.

Subsubsection Examples

Example 1.52.

Solve the equation \(2x-3y=8\) for \(y\text{.}\)

Solution
\begin{align*} 2x-3y \amp = 8 \amp \amp \blert{\text{Subtract from both sides.}} \\ -3y \amp = 8-2x \amp \amp \blert{\text{Divide both sides by} -3.}\\ y \amp =\dfrac{8-2x}{-3} \amp \amp \blert{\text{Divide each term of the numerator by} -3.}\\ y \amp =\dfrac{8}{3}-\dfrac{2}{3}x \end{align*}
Example 1.53.

Solve the equation \(A=\dfrac{h}{2}(b+c)\) for \(b\text{.}\)

Solution

It is nearly always best to clear fractions from an equation first, so we begin by multiplying both sides by 2.

\begin{align*} \alert{2}A \amp = \alert{\cancel{2}}\left(\dfrac{h}{\cancel{2}}(b+c)\right) \amp \amp \blert{\text{Multiply both sides by}~ \alert{2}.}\\ 2A \amp = h(b+c) \amp \amp \blert{\text{Divide both sides by}~h.}\\ \dfrac{2A}{h} \amp = b+c \amp \amp \blert{\text{Subtract from both sides.}}\\ \dfrac{2A}{h}-c \amp = b \end{align*}

Subsubsection Exercises

Solve \(f=s+at\) for \(t\)

Answer
\(t=\dfrac{f-a}{s}\)

Solve \(2x-4y=k\) for \(y\)

Answer
\(y=\dfrac{-1}{4}k+\dfrac{1}{2}x\)

Solve \(P=2l+2w\) for \(l\)

Answer
\(l=\dfrac{P}{2}-w\)

Solve \(\dfrac{x}{a}+\dfrac{y}{b}=1\) for \(x\)

Answer
\(x=a-\dfrac{ay}{b}\)