# 4.3: Graphing Using Intercepts

**At Grade**Created by: CK-12

## Learning Objectives

- Find intercepts of the graph of an equation.
- Use intercepts to graph an equation.
- Solve real-world problems using intercepts of a graph

## Introduction

Sanjit’s office is 25 miles from home, and in traffic he expects the trip home to take him an hour if he starts at 5 PM. Today he hopes to stop at the post office along the way. If the post office is 6 miles from his office, when will Sanjit get there?

If you know just one of the points on a line, you’ll find that isn’t enough information to plot the line on a graph. As you can see in the graph above, there are many lines—in fact, infinitely many lines—that pass through a single point. But what if you know two points that are both on the line? Then there’s only one way to graph that line; all you need to do is plot the two points and use a ruler to draw the line that passes through both of them.

There are a lot of options for choosing which two points on the line you use to plot it. In this lesson, we’ll focus on two points that are rather convenient for graphing: the points where our line crosses the \begin{align*}x- \end{align*}**intercepts.** We’ll see how to find intercepts algebraically and use them to quickly plot graphs.

Look at the graph above. The **\begin{align*}y-\end{align*} y−intercept** occurs at the point where the graph crosses the \begin{align*}y-\end{align*}

Similarly, the **\begin{align*}x-\end{align*}intercept** occurs at the point where the graph crosses the \begin{align*}x-\end{align*}axis. The \begin{align*}x-\end{align*}value at this point is 6, and the \begin{align*}y-\end{align*}value is 0.

So we know the coordinates of two points on the graph: (0, 8) and (6, 0). If we’d just been given those two coordinates out of the blue, we could quickly plot those points and join them with a line to recreate the above graph.

**Note:** Not all lines will have both an \begin{align*}x-\end{align*}and a \begin{align*}y-\end{align*}intercept, but most do. However, horizontal lines never cross the \begin{align*}x-\end{align*}axis and vertical lines never cross the \begin{align*}y-\end{align*}axis.

For examples of these special cases, see the graph below.

## Finding Intercepts by Substitution

**Example 1**

*Find the intercepts of the line \begin{align*}y=13-x\end{align*} and use them to graph the function.*

**Solution**

The first intercept is easy to find. The \begin{align*}y-\end{align*}intercept occurs when \begin{align*}x=0\end{align*}. Substituting gives us \begin{align*}y=13-0=13\end{align*}, so the \begin{align*}y-\end{align*}intercept is (0, 13).

Similarly, the \begin{align*}x-\end{align*}intercept occurs when \begin{align*}y = 0\end{align*}. Plugging in 0 for \begin{align*}y\end{align*} gives us \begin{align*}0=13-x\end{align*}, and adding \begin{align*}x\end{align*} to both sides gives us \begin{align*}x=13\end{align*}. So (13, 0) is the \begin{align*}x-\end{align*}intercept.

To draw the graph, simply plot these points and join them with a line.

**Example 2**

*Graph the following functions by finding intercepts.*

a) \begin{align*} y=2x+3\end{align*}

b) \begin{align*} y=7-2x\end{align*}

c) \begin{align*} 4x-2y=8\end{align*}

d) \begin{align*}2x+3y=-6\end{align*}

**Solution**

a) Find the \begin{align*}y-\end{align*}intercept by plugging in \begin{align*}x = 0:\end{align*}

\begin{align*}y = 2 \cdot 0 + 3 = 3 \qquad - \text{the} \ y-\text{intercept is} \ (0, 3)\end{align*}

Find the \begin{align*}x-\end{align*}intercept by plugging in \begin{align*}y = 0:\end{align*}

\begin{align*}0 & = 2x + 3 \qquad - subtract \ 3 \ from \ both \ sides:\\ -3 & = 2x \qquad \quad \ \ - divide \ by \ 2:\\ -\frac {3}{2} & = x \qquad \qquad - \text{the} \ x-\text{intercept is} \ (-1.5, 0)\end{align*}

b) Find the \begin{align*}y-\end{align*}intercept by plugging in \begin{align*}x = 0:\end{align*}

\begin{align*}y = 7 - 2 \cdot 0 = 7 \qquad - \text{the} \ y-\text{intercept is} \ (0, 7)\end{align*}

Find the \begin{align*}x-\end{align*}intercept by plugging in \begin{align*}y = 0:\end{align*}

\begin{align*}0 &= 7 - 2x \qquad - subtract \ 7 \ from \ both \ sides:\\ -7 & = -2x \qquad \ \ \ - divide \ by \ -2:\\ \frac {7} {2} & = x \qquad \qquad \ - \text{the} \ x-\text{intercept is} \ (3.5, 0)\end{align*}

c) Find the \begin{align*}y-\end{align*}intercept by plugging in \begin{align*}x = 0:\end{align*}

\begin{align*}4 \cdot 0 - 2y & = 8\\ -2y & = 8 \qquad \ \ - divide \ by \ - 2\\ y & = - 4 \qquad - \text{the} \ y-\text{intercept is} \ (0, -4)\end{align*}

Find the \begin{align*}x-\end{align*}intercept by plugging in \begin{align*}y = 0:\end{align*}

\begin{align*}4x - 2\cdot 0 & = 8\\ 4x & = 8 \qquad - divide \ by \ 4:\\ x & = 2 \qquad - \text{the} \ x-\text{intercept is} \ (2, 0)\end{align*}

d) Find the \begin{align*}y-\end{align*}intercept by plugging in \begin{align*}x = 0:\end{align*}

\begin{align*}2 \cdot 0 + 3y & = - 6\\ 3y & = - 6 \qquad - \text{divide by} \ 3:\\ y & = - 2 \qquad - \text{the} \ y-\text{intercept is} \ (0, - 2)\end{align*}

Find the \begin{align*}x-\end{align*}intercept by plugging in \begin{align*}y = 0:\end{align*}

\begin{align*}2x + 3 \cdot 0 & = -6\\ 2x & = - 6 \qquad - \text{divide by} \ 2:\\ x & = - 3 \qquad - \text{the} \ x-\text{intercept is} \ (- 3, 0)\end{align*}

## Finding Intercepts for Standard Form Equations Using the Cover-Up Method

Look at the last two equations in example 2. These equations are written in ** standard form.** Standard form equations are always written “

**coefficient**times \begin{align*}x\end{align*} plus (or minus)

**coefficient**times \begin{align*}y\end{align*} equals

**value**”. In other words, they look like this:

\begin{align*}ax+by=c\end{align*}

where \begin{align*}a\end{align*} has to be positive, but \begin{align*}b\end{align*} and \begin{align*}c\end{align*} do not.

There is a neat method for finding intercepts in standard form, often referred to as the cover-up method.

**Example 3**

*Find the intercepts of the following equations:*

a) \begin{align*}7x - 3y = 21\end{align*}

b) \begin{align*}12x - 10y = -15\end{align*}

c) \begin{align*}x + 3y = 6\end{align*}

**Solution**

To solve for each intercept, we realize that at the intercepts the value of **either** \begin{align*}x\end{align*} **or** \begin{align*}y\end{align*} is zero, and so any terms that contain that variable effectively drop out of the equation. To make a term disappear, simply cover it (a finger is an excellent way to cover up terms) and solve the resulting equation.

a) To solve for the \begin{align*}y-\end{align*}intercept we set \begin{align*}x = 0\end{align*} and cover up the \begin{align*}x-\end{align*}term:

\begin{align*}- 3y = 21\! \\ y = -7 \qquad (0, -7) \ \text{is the} \ y-\text{intercept.}\end{align*}

Now we solve for the \begin{align*}x-\end{align*}intercept:

\begin{align*}7x = 21\! \\ x = 3 \qquad (3, 0) \ \text{is the} \ x-\text{intercept.}\end{align*}

b) To solve for the \begin{align*}y-\end{align*}intercept \begin{align*}(x = 0)\end{align*}, cover up the \begin{align*}x-\end{align*}term:

\begin{align*}-10y = - 15\! \\ y = 1.5 \qquad (0, 1.5) \ \text{is the} \ y-\text{intercept.}\end{align*}

Now solve for the \begin{align*}x-\end{align*}intercept \begin{align*}(y = 0)\end{align*}:

\begin{align*}12x = - 15\! \\ x = - \frac{5}{4} \qquad (-1.25, 0) \ \text{is the} \ x-\text{intercept.}\end{align*}

c) To solve for the \begin{align*}y-\end{align*}intercept \begin{align*}(x = 0)\end{align*}, cover up the \begin{align*}x-\end{align*}term:

\begin{align*}3y = 6\! \\ y = 2 \qquad (0, 2) \ \text{is the} \ y-\text{intercept.}\end{align*}

Solve for the \begin{align*}y-\end{align*}intercept:

\begin{align*}x = 6 \qquad (6, 0) \ \text{is the} \ x-\text{intercept.}\end{align*}

The graph of these functions and the intercepts is below:

To learn more about equations in standard form, try the Java applet at http://www.analyzemath.com/line/line.htm (scroll down and click the “click here to start” button.) You can use the sliders to change the values of \begin{align*}a, b,\end{align*} and \begin{align*}c\end{align*} and see how that affects the graph.

## Solving Real-World Problems Using Intercepts of a Graph

**Example 4**

*Jesus has $30 to spend on food for a class barbecue. Hot dogs cost $0.75 each (including the bun) and burgers cost $1.25 (including the bun). Plot a graph that shows all the combinations of hot dogs and burgers he could buy for the barbecue, without spending more than $30.*

This time we will find an equation first, and then we can think logically about finding the intercepts.

If the number of burgers that Jesus buys is \begin{align*}x\end{align*}, then the money he spends on burgers is \begin{align*}1.25x\end{align*}

If the number of hot dogs he buys is \begin{align*}y\end{align*}, then the money he spends on hot dogs is \begin{align*}0.75y\end{align*}

So the total cost of the food is \begin{align*}1.25x + 0.75y\end{align*}.

The total amount of money he has to spend is $30, so if he is to spend it ALL, we can use the following equation:

\begin{align*}1.25x + 0.75y = 30\end{align*}

We can solve for the intercepts using the cover-up method. First the \begin{align*}y-\end{align*}intercept \begin{align*}(x = 0)\end{align*}:

\begin{align*}0.75y = 30\! \\ y = 40 \qquad y-\text{intercept:} \ (0, 40)\end{align*}

Then the \begin{align*}x-\end{align*}intercept \begin{align*}(y = 0)\end{align*}:

\begin{align*}1.25x = 30\! \\ x = 24 \qquad x-\text{intercept:} \ (24, 0)\end{align*}

Now we plot those two points and join them to create our graph, shown here:

We could also have created this graph without needing to come up with an equation. We know that if John were to spend ALL the money on hot dogs, he could buy \begin{align*}\frac{30}{.75}=40\end{align*} hot dogs. And if he were to buy only burgers he could buy \begin{align*}\frac{30}{1.25}=24\end{align*} burgers. From those numbers, we can get 2 intercepts: (0 burgers, 40 hot dogs) and (24 burgers, 0 hot dogs). We could plot these just as we did above and obtain our graph that way.

As a final note, we should realize that Jesus’ problem is really an example of an inequality. He can, in fact, spend any amount up to $30. The only thing he cannot do is spend more than $30. The graph above reflects this: the line is the set of solutions that involve spending exactly $30, and the shaded region shows solutions that involve spending less than $30. We’ll work with inequalities some more in Chapter 6.

## Lesson Summary

- A \begin{align*}y-\end{align*}
**intercept**occurs at the point where a graph crosses the \begin{align*}y-\end{align*}axis (where \begin{align*}x=0\end{align*}) and an \begin{align*}x-\end{align*}**intercept**occurs at the point where a graph crosses the \begin{align*}x-\end{align*}axis (where \begin{align*}y=0\end{align*}). - The \begin{align*}y-\end{align*}intercept can be found by
**substituting**\begin{align*}x=0\end{align*} into the equation and solving for \begin{align*}y\end{align*}. Likewise, the \begin{align*}x-\end{align*}intercept can be found by**substituting**\begin{align*} y=0\end{align*} into the equation and solving for \begin{align*}x\end{align*}. - A linear equation is in
**standard form**if it is written as “positive coefficient times \begin{align*}x\end{align*} plus coefficient times \begin{align*}y\end{align*} equals value”. Equations in standard form can be solved for the intercepts by covering up the \begin{align*}x\end{align*} (or \begin{align*}y\end{align*}) term and solving the equation that remains.

## Review Questions

- Find the intercepts for the following equations using substitution.
- \begin{align*}y=3x-6\end{align*}
- \begin{align*}y=-2x+4\end{align*}
- \begin{align*}y=14x-21\end{align*}
- \begin{align*}y=7-3x\end{align*}
- \begin{align*}y=2.5x-4\end{align*}
- \begin{align*}y=1.1x+2.2\end{align*}
- \begin{align*}y= \frac {3} {8} x+7\end{align*}
- \begin{align*}y=\frac {5} {9} - \frac {2} {7} x\end{align*}

- Find the intercepts of the following equations using the cover-up method.
- \begin{align*}5x-6y=15\end{align*}
- \begin{align*}3x-4y=-5\end{align*}
- \begin{align*}2x+7y=-11\end{align*}
- \begin{align*}5x+10y=25\end{align*}
- \begin{align*}5x-1.3y=12\end{align*}
- \begin{align*}1.4x -3.5y=7\end{align*}
- \begin{align*} \frac {3}{5} x + 2y = \frac {2}{5}\end{align*}
- \begin{align*}\frac {3}{4} x - \frac{2}{3}y = \frac {1}{5}\end{align*}

- Use any method to find the intercepts and then graph the following equations.
- \begin{align*}y=2x+3\end{align*}
- \begin{align*}6(x-1) = 2(y+3)\end{align*}
- \begin{align*}x-y=5\end{align*}
- \begin{align*}x+y=8\end{align*}

- At the local grocery store strawberries cost $3.00 per pound and bananas cost $1.00 per pound.
- If I have $10 to spend on strawberries and bananas, draw a graph to show what combinations of each I can buy and spend exactly $10.
- Plot the point representing 3 pounds of strawberries and 2 pounds of bananas. Will that cost more or less than $10?
- Do the same for the point representing 1 pound of strawberries and 5 pounds of bananas.

- A movie theater charges $7.50 for adult tickets and $4.50 for children. If the theater takes in $900 in ticket sales for a particular screening, draw a graph which depicts the possibilities for the number of adult tickets and the number of child tickets sold.
- Why can't we use the intercept method to graph the following equation? \begin{align*}3(x+2) =2 (y+3)\end{align*}
- Name two more equations that we can’t use the intercept method to graph.

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