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Section 3.1 Intercepts

Subsection Intercepts of a Line

The intercepts of a line are the points where the graph crosses the axes.

It is easy to recognize the intercepts of a line on a good graph. Here is a graph of the line

\begin{equation*} y=\dfrac{1}{2}x+4 \end{equation*}

We can see that its \(x\)-intercept is the point \((-8,0)\text{,}\) and its \(y\)-intercept is \((0,4)\text{.}\)

If we have an equation for a line we can find the intercepts of its graph algebraically. In Example 1 we'll use algebra to find the intercepts of this line.

graph of line
Look Closer.

Look at the intercepts on the graph above. Because the \(y\)-intercept of a graph lies on the \(y\)-axis, its \(x\)-coordinate must be zero. And because the \(x\)-intercept lies on the \(x\)-axis, its \(y\)-coordinate must be zero.

Example 3.1.

Use algebra to find the intercepts of the graph of \(~y=\dfrac{1}{2}x+4\text{.}\)

Solution

To find the \(y\)-intercept, we substitute \(\alert{0}\) for \(x\) in the equation and solve for \(y\text{.}\)

\begin{equation*} y=\dfrac{1}{2}(\alert{0})+4 = 4 \end{equation*}

The \(y\)-intercept is \((0,4)\text{,}\) as we saw above. To find the \(x\)-intercept, we substitute \(\alert{0}\) for \(y\) in the equation and solve for \(x\text{.}\)

\begin{equation*} \begin{aligned} \alert{0} \amp = \dfrac{1}{2}x+4 \amp \amp \blert{\text{Subtract 4 from both sides.}}\\ -4 \amp = \dfrac{1}{2}x \amp \amp \blert{\text{Multiply both sides by 2.}}\\ -8 \amp = x \end{aligned} \end{equation*}

The \(x\)-intercept is \((-8,0)\text{,}\) as expected.

To find the intercepts of a graph.
  • To find the \(x\)-intercept of a graph:

    Substitute \(0\) for \(y\) in the equation and solve for \(x\text{.}\)

  • To find the \(y\)-intercept of a graph:

    Substitute \(0\) for \(x\) in the equation and solve for \(y\text{.}\)

Reading Questions Reading Questions

1.

What are the intercepts of a line?

2.

How do we find the \(x\)-intercept of the graph of an equation?

Subsection The Intercept Method of Graphing

We can use the \(x\)- and \(y\)-intercepts to graph a linear equation quickly. Instead of choosing several different values of \(x\) to find points on the graph, we find the two intercepts and fill in the short table shown below. By finding the missing values for \(x\) and \(y\) we are finding the intercepts of the graph.

\(x\) \(y\)
\(0\) \(\hphantom{0000}\)
\(\hphantom{0000}\) \(0\)

Example 3.2.

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 of the graph, we substitute \(0\) for \(x\) and solve for \(y\text{.}\)

\begin{equation*} \begin{aligned} 3(\alert{0}) + 2y \amp = 7 \amp \amp \blert{\text{Simplify 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{aligned} \end{equation*}

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

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

\begin{equation*} \begin{aligned} 3 + 2(\alert{0}) \amp = 7 \amp \amp \blert{\text{Simplify 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{aligned} \end{equation*}

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

Here is a table showing the two intercepts. We plot the intercepts and connect them with a straight line to obtain the graph below.

\(x\) \(y\)
\(0\) \(3\dfrac{1}{2}\)
\(2\dfrac{1}{3} \) \(0\)
graph of line and intercepts

It is a good idea to find a third point as a check. We choose \(x=\alert{1}\) and solve for \(y\text{.}\)

\begin{equation*} \begin{aligned} 3 + 2(\alert{1}) \amp = 7 \amp \amp \blert{\text{Subtract 3 from both sides.}}\\ 2y \amp = 4 \amp \amp \blert{\text{Divide both sides by 2.}}\\ y \amp = 2 \end{aligned} \end{equation*}

You can check that the point \((1,2)\) lies on the graph.

The intercept method for graphing a line is often faster and more efficient than making a table and plotting points.

To Graph a Linear Equation Using the Intercept Method.
  1. Find the \(x\)- and \(y\)-intercepts of the graph.

  2. Draw the line through the two intercepts.
  3. Find a third point on the graph as a check. (Choose any convenient value for \(x\) and solve for \(y\text{.}\))

Reading Questions Reading Questions

3.

How do we graph a line by the intercept method?

4.

Make a table that you can use with the intercept method.

5.

What is the \(y\)-intercept of the line \(3x+2y=7\text{?}\)

Subsection Interpreting the Intercepts

The intercepts of a graph give us valuable information about a problem. They often represent the starting or ending values for a particular variable.

Example 3.3.

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

  1. Write and graph an equation for the temperature \(T\) at \(h\) hours after noon.
  2. Find the intercepts of the graph. What do the intercepts tell us about the temperature in Nome?
Solution
  1. An equation for the temperature is

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

    A graph of the equation is shown below.

    graph: line and intercepts

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

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

    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{equation*} \begin{aligned} \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{aligned} \end{equation*}

    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.

Reading Questions Reading Questions

6.

The intercepts of a graph often represent the or values for a particular variable.

Subsection Skills Warm-Up

Exercises Exercises

Choose the correct algebraic expression for each of the following situations.

\begin{gather*} 5x-8=30 ~~~~~~~~ \dfrac{x}{5}-30=8 ~~~~~~~~ \dfrac{x-30}{5}=8\\ 5x+8=30 ~~~~~~~~ \dfrac{x}{5}+30=8 ~~~~~~~~ \dfrac{x+30}{5}=8 \end{gather*}
1.

Ilciar has earned a total of 30 points on the first four quizzes in his biology class. What must he earn on the fifth quiz to end up with an average of 8?

2.

Jocelyn ordered five exotic plants from a nursery. She paid a total of $30, including an $8 shipping fee. How much did she pay for each plant?

3.

To buy new equipment, the five members of the chess club used $30 from the treasury, and each member donated $8. How much did the equipment cost?

4.

Hemman bought 5 tapes on sale, and he cashed in a gift certificate for $8. He then owed the clerk $30. How much was each tape?

5.

Nirusha and four other people won the office baseball pool. After spending $30 of her share, Nirusha had $8 left. What was the total amount in the pool?

Solutions Answers to Skills Warm-Up

Exercises Exercises

Exercises Homework 3.1

For Problems 1–6,

  1. Find the \(x\)- and \(y\)-intercepts of the line.
  2. Use the intercept method to graph the line.
1.
\(2x+4y=8\)
grid
2.
\(x+2y+10=0\)
grid
3.
\(2x=14+7y\)
grid
4.
\(y=-4x+8\)
grid
5.
\(\dfrac{x}{20}+\dfrac{y}{30}=1\)
50 by 50 grid
6.
\(3x-2y=120\)
100 by 100 grid

For Problems 7–12, match each equation with its graph. (More than one equation may describe the same graph.)

line
line
line
line
line
line
7.
\(2x+3y=12\)
8.
\(2x-3y=12\)
9.
\(3x-2y=12\)
10.
\(-3x-2y=12\)
11.
\(\dfrac{x}{6}-\dfrac{y}{4}=1\)
12.
\(\dfrac{x}{4}-\dfrac{y}{6}=1\)

For Problems 13–18, solve each pair of equations. In part (b) of each problem, your answer will involve the constant \(k\text{.}\)

13.
  1. \(-2x=6\)
  2. \(-2x=k\)
14.
  1. \(x+4=7\)
  2. \(x+4=k\)
15.
  1. \(x-5=9\)
  2. \(x-5=k\)
16.
  1. \(2x+3=8\)
  2. \(2x+k=8\)
17.
  1. \(15-4x=3\)
  2. \(15-4x=k\)
18.
  1. \(9+3x=-1\)
  2. \(9+kx=-1\)
19.

During spring break, Francine took the train to San Francisco and then bicycled home. The graph below left shows Francine's distance \(d\) from home, in miles, after cycling for \(h\) hours. Use the graph to answer the questions.

graph of line

  1. How far is it from San Francisco to Francine's home?
  2. How many hours did Francine cycle to get home?
  3. After cycling for 12 hours, how far is Francine from home?
  4. How far did Francine cycle in the first four hours?

For Problems 20–22,

  1. Find the intercepts of each linear equation.
  2. Use the intercept method to graph the line.
  3. Explain what the intercepts mean in terms of the problem situation.
20.

The amount of home heating oil (in gallons) in the Olsons' tank is given by the equation \(G=200-15w\text{,}\) where \(w\) is the number of weeks since they turned on the furnace.

grid
21.

Dana joined a savings plan some weeks ago. Her bank balance is growing each week according to the formula \(B=225+25w\text{,}\) where \(w=0\) represents this week.

grid
22.

Delbert bought some equipment and went into the dog-grooming business. His profit is increasing according to the equation \(P=-600+40d\text{,}\) where \(d\) is the number of dogs he has groomed.

grid
23.
  1. At what point does the graph of \(2x-3y=35\) cross the \(x\)-axis?

  2. At what point does the graph of \(1.4x+3.6y=-18\) cross the \(y\)-axis?

24.

The \(x\)-intercept of a line is positive and its \(y\)-intercept is negative. Is the line increasing or decreasing? Sketch a possible example of such a line.