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4.3: Graphing Using Intercepts

Difficulty Level: 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

Only two distinct points are needed to uniquely define a graph of a line. After all, there are an infinite number of lines that pass through a single point (a few are shown in the graph at right). But if you supplied just one more point, there can only be one line that passes through both points. To plot the line, just plot the two points and use a ruler, edge placed on both points, to trace the graph of the line.

There are a lot of options for choosing which two points on the line you use to plot it. In this lesson, we will focus on two points that are rather convenient for graphing: the points where our line crosses the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} axes, or intercepts. We will be finding intercepts algebraically and using them to quickly plot graphs. 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 in the graph at the right is 6.

Look at the graph to the right. The \begin{align*}y-\end{align*}intercept occurs at the point where the graph crosses the \begin{align*}y-\end{align*}axis. The \begin{align*}y-\end{align*}value at this point is 8.

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.

Now we know that the \begin{align*}x\end{align*} value of all the points on the \begin{align*}y-\end{align*}axis is zero, and the \begin{align*}y\end{align*} value of all the points on the \begin{align*}x-\end{align*}axis is also zero. So if we were given the coordinates of the two intercepts (0, 8) and (6, 0) we could quickly plot these points and join them with a line to recreate our graph.

Note: Not all lines will have both intercepts but most do. Specifically, 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 case lines, see the graph at right.

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.

The first intercept is easy to find. The \begin{align*}y-\end{align*}intercept occurs when \begin{align*}x=0\end{align*} Substituting gives:

\begin{align*}y = 13 - 0 = 13 && (0, 13)\ \text{is the}\ x-\text{intercept.}\end{align*}

We know that the \begin{align*}x-\end{align*}intercept has, by definition, a \begin{align*}y-\end{align*}value of zero. Finding the corresponding \begin{align*}x-\end{align*}value is a simple case of substitution:

\begin{align*}0 & = 13 - x && \text{To isolate}\ x\ \text{subtract}\ 13\ \text{from both sides}.\\ -13 & = -x && \text{Divide by} -1.\end{align*}

Solution

(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*}

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 && \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 && \text{Subtract}\ 3\ \text{from both sides.}\\ -3 & = 2x && \text{Divide by}\ 2. \\ - \frac{3}{2}& = x && \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 && \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 && \text{Subtract}\ 7\ \text{from both sides.} \\ -7 & = -2x && \text{Divide by}\ -2.\\ \frac{7} {2} & = x && \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 && \text{Divide by}\ -2.\\ y & = -4 && \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 && \text{Divide by}\ 4. \\ x & = 2 && \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 && \text{Divide by}\ 3.\\ y & = -2 && \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 && \text{Divide by}\ 2.\\ x &=-3 && \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 “positive coefficient times \begin{align*}x\end{align*} plus (or minus) positive coefficient times \begin{align*}y\end{align*} equals value”. Note that the \begin{align*}x\end{align*} term always has a positive value in front of it while the \begin{align*}y\end{align*} value may have a negative term. The equation looks like this:

\begin{align*}ax + by + c\ \text{or}\ ax - by = c && (a\ \text{and}\ b\ \text{are positive numbers})\end{align*}

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*}

To solve for each intercept, we realize that on 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 the zero variable effectively disappear. 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 \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 \qquad (3, 0)\ \text{is the}\ x-\text{intercept}.\end{align*}

b. Solve for the \begin{align*}y-\end{align*}intercept \begin{align*}(x = 0)\end{align*} by covering up the \begin{align*}x\end{align*} term.

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

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

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

c. Solve for the \begin{align*}y-\end{align*}intercept \begin{align*}(x = 0)\end{align*} by covering up the \begin{align*}x\end{align*} term:

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

The graph of these functions and the intercepts is shown in the graph on the right.

Solving Real-World Problems Using Intercepts of a Graph

Example 4

The monthly membership cost of a gym is $25 per month. To attract members, the gym is offering a$100 cash rebate if members sign up for a full year. Plot the cost of gym membership over a 12 month period. Use the graph to determine the final cost for a 12 month membership.

Let us examine the problem. Clearly the cost is a function of the number of months (not the other way around). Our independent variable is the number of months (the domain will be whole numbers) and this will be our \begin{align*}x\end{align*} value. The cost in dollars is the dependent variable and will be our \begin{align*}y\end{align*} value. Every month that passes the money paid to the gym goes up by $25. However, we start with a$100 cash gift, so our initial cost (\begin{align*}y-\end{align*}intercept) is 100. This pays for four months \begin{align*}(4 \times \25 = 100)\end{align*} so after four months the cost of membership (\begin{align*}y-\end{align*}value) is zero. The \begin{align*}y-\end{align*}intercept is (0, -100). The \begin{align*}x-\end{align*}intercept is (4, 0). We plot our points, join them with a straight line and extend that line out all the way to the \begin{align*}x = 12\end{align*} line. The graph is shown below. To find the cost of a 12 month membership we simply read off the value of the function at the 12 month point. A line drawn up from \begin{align*}x = 12\end{align*} on the \begin{align*}x\end{align*} axis meets the function at a \begin{align*}y\end{align*} value of200.

Solution

The cost of joining the gym for one year is $200. Example 5 Jesus has$30 to spend on food for a class barbeque. Hot dogs cost $0.75 each (including the bun) and burgers cost$1.25 (including bun and salad). 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 John buys is \begin{align*}x\end{align*}, then the money spent 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 spent on hot dogs is \begin{align*}0.75y\end{align*}. \begin{align*}1.25x + 0.75y && \text{The total cost of the food.}\end{align*} The total amount of money he has to spend is30. If he is to spend it ALL, then we can use the following equation.

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

We 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 \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 \qquad x-\text{intercept} \ (24, 0)\end{align*}

We can now plot the points and join them to create our graph, shown right.

Here is an alternative to the equation method.

If Jesus were to spend ALL the money on hot dogs, he could buy \begin{align*} \frac{30} {0.75} = 40\end{align*} hot dogs. If on the other hand, he were to buy only burgers, he could buy \begin{align*} \frac{30} {1.25} = 24\end{align*} burgers. So you can see that we get two intercepts: (0 burgers, 40 hot dogs) and (24 burgers, 0 hot dogs). We would plot these in an identical manner and design 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. So our graph reflects this. The shaded region shows where Jesus' solutions all lie. We will see inequalities again 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 \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 \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 (or minus) positive 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

1. Find the intercepts for the following equations using substitution.
1. \begin{align*}y = 3x - 6\end{align*}
2. \begin{align*}y = -2x + 4 \end{align*}
3. \begin{align*}y = 14x- 21 \end{align*}
4. \begin{align*}y = 7 -3x\end{align*}
2. Find the intercepts of the following equations using the cover-up method.
1. \begin{align*}5x- 6y = 15\end{align*}
2. \begin{align*}3x-4y = -5\end{align*}
3. \begin{align*}2x+7y = -11 \end{align*}
4. \begin{align*}5x + 10y = 25\end{align*}
3. Use any method to find the intercepts and then graph the following equations.
1. \begin{align*}y = 2x + 3 \end{align*}
2. \begin{align*}6(x- 1) = 2(y + 3) \end{align*}
3. \begin{align*}x-y = 5 \end{align*}
4. \begin{align*}x+y = 8\end{align*}
4. At the local grocery store strawberries cost $3.00 per pound and bananas cost$1.00 per pound. If I have $10 to spend between strawberries and bananas, draw a graph to show what combinations of each I can buy and spend exactly$10.
5. A movie theater charges $7.50 for adult tickets and$4.50 for children. If the theater takes \$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.
6. Why can't we use the intercept method to graph the following equation? \begin{align*}3(x + 2) = 2(y+3)\end{align*}

1. (0, -6), (2, 0)
2. (0, 4), (2, 0)
3. (0, -21), (1.5, 0)
4. (0, 7), \begin{align*}\left (\frac{7}{3}, 0\right )\end{align*}
1. (0, -2.5), (3, 0)
2. (0, 1.25), \begin{align*}\left (-\frac{5}{3}, 0\right )\end{align*}
3. \begin{align*}\left (0,-\frac{11}{7}\right ), \left (-\frac{11}{2},0\right ) \end{align*}
4. (0, 2.5), (5, 0)
1. This equation reduces to \begin{align*}3x=2y\end{align*}, which passes through (0, 0) and therefore only has one intercept. Two intercepts are needed for this method to work.

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