**Standard**

**MCC9-12.F.IF.7a Graph linear functions and show intercepts.**^{ }

Suppose the linear function *f(x) = -0.25x + 10* represents the amount of money you have left to play video games, where *f(x)* is the amount of money you have left and x is the number of video games that you have played so far. Do you know how to graph this function? What would be the slope and intercept of the graph? In this Concept, you'll learn how to graph linear functions like this one by finding the graph's slope and intercept.

### Guidance

You can see that the notations are interchangeable. This means you can substitute the notation for and use all the concepts you have learned on linear equations.

#### Example A

\begin{align*}\text{Graph} \ f(x)& =\frac{1}{3}x+1.\\ \text{Replace} \ f(x)& = \text{with} \ y=.\\ y& =\frac{1}{3} x+1\end{align*}

This equation is in slope-intercept form. You can now graph the function by graphing the intercept and then using the slope as a set of directions to find your second coordinate.

### Watch This

CK-12 Foundation: 0408S Graphs Using Slope-Intercept Form (H264)

### Try This

To get a better understanding of what happens when you change the slope or the \begin{align*}y-\end{align*}intercept of a linear equation, try playing with the Java applet at http://standards.nctm.org/document/eexamples/chap7/7.5/index.htm.

**Identify Slope and \begin{align*}y-\end{align*}intercept**

So far, we’ve been writing a lot of our equations in **slope-intercept form—**that is, we’ve been writing them in the form \begin{align*}y = mx + b\end{align*}, where \begin{align*}m\end{align*} and \begin{align*}b\end{align*} are both constants. It just so happens that \begin{align*}m\end{align*} is the slope and the point \begin{align*}(0, b)\end{align*} is the \begin{align*}y-\end{align*}intercept of the graph of the equation, which gives us enough information to draw the graph quickly.

#### Example A

*Identify the slope and \begin{align*}y-\end{align*}intercept of the following equations.*

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

b) \begin{align*}y = 0.5x - 3\end{align*}

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

d) \begin{align*}y = -4\end{align*}

**Solution**

a) Comparing , we can see that \begin{align*}m = 3\end{align*} and \begin{align*}b = 2\end{align*}. So \begin{align*}y = 3x + 2\end{align*} has a **slope of 3** and a \begin{align*}y-\end{align*}**intercept of (0, 2).**

b) has a **slope of 0.5** and a \begin{align*}y-\end{align*}**intercept of (0, -3).**

Notice that the intercept is **negative**. The \begin{align*}b-\end{align*}term includes the sign of the operator (plus or minus) in front of the number—for example, \begin{align*}y = 0.5x - 3\end{align*} is identical to \begin{align*}y = 0.5x + (-3)\end{align*}, and that means that \begin{align*}b\end{align*} is -3, not just 3.

c) At first glance, this equation doesn’t look like it’s in slope-intercept form. But we can rewrite it as \begin{align*}y = -7x + 0\end{align*}, and that means it has a **slope of -7** and a \begin{align*}y-\end{align*}**intercept of (0, 0).** Notice that the slope is negative and the line passes through the origin.

d) We can rewrite this one as \begin{align*}y = 0x - 4\end{align*}, giving us a **slope of 0** and a \begin{align*}y-\end{align*}**intercept of (0, -4).** This is a horizontal line.

** Graph an Equation in Slope-Intercept Form**

Once we know the slope and intercept of a line, it’s easy to graph it.

*Ali is trying to work out a trick that his friend showed him. His friend started by asking him to think of a number, then double it, then add five to the result. Ali has written down a rule to describe the first part of the trick. He is using the letter \begin{align*}x\end{align*} to stand for the number he thought of and the letter \begin{align*}y\end{align*} to represent the final result of applying the rule. He wrote his rule in the form of an equation: \begin{align*}y = 2x + 5\end{align*}.*

*Help him visualize what is going on by graphing the function that this rule describes.*

In that example, we constructed a table of values, and used that table to plot some points to create our graph.

We also saw another way to graph this equation. Just by looking at the equation, we could see that the \begin{align*}y-\end{align*}intercept was (0, 5), so we could start by plotting that point. Then we could also see that the slope was 2, so we could find another point on the graph by going up 2 units and right one unit. The graph would then be the line between those two points.

Here’s another problem where we can use the same method.

#### Example C

*Graph the following function: \begin{align*}y=-3x+5\end{align*}*

**Solution**

To graph the function without making a table, follow these steps:

- Identify the \begin{align*}y-\end{align*}intercept: \begin{align*}b = 5\end{align*}
- Plot the intercept: (0, 5)
- Identify the slope: \begin{align*}m = -3\end{align*}. (This is equal to \begin{align*}\frac {-3}{1}\end{align*}, so the
**rise**is -3 and the**run**is 1.) - Move
**down**3 units and**right**one unit to find another point on the line: (1, 2) - Draw the line through the points (0, 5) and (1, 2).

Notice that to graph this equation based on its slope, we had to find the rise and run—and it was easiest to do that when the slope was expressed as a fraction. That’s true in general: to graph a line with a particular slope, it’s easiest to first express the slope as a fraction in simplest form, and then read off the numerator and the denominator of the fraction to get the rise and run of the graph.

#### Example D

*Find integer values for the* *rise**and* *run**of the following slopes, then graph lines with corresponding slopes.*

a) \begin{align*}m=3\end{align*}

b) \begin{align*}m=-2\end{align*}

**Solution**

a)

b)

### Vocabulary

- A common form of a line (linear equation) is
**slope-intercept form:**\begin{align*}y=mx+b\end{align*}, where \begin{align*}m\end{align*} is the slope and the point \begin{align*}(0, b)\end{align*} is the \begin{align*}y-\end{align*}intercept - Graphing a line in slope-intercept form is a matter of first plotting the \begin{align*}y-\end{align*}intercept \begin{align*}(0, b)\end{align*}, then finding a second point based on the slope, and using those two points to graph the line.

### Guided Practice

*Find integer values for the* *rise**and* *run**of the following slopes, then graph lines with corresponding slopes.*

a) \begin{align*}m=0.75\end{align*}

b) \begin{align*}m=-0.375\end{align*}

**Solution:**

a)

b)

### Practice

Identify the slope and \begin{align*}y-\end{align*}intercept for the following equations.

- \begin{align*}y=2x+5\end{align*}
- \begin{align*}y=-0.2x+7\end{align*}
- \begin{align*}y=x\end{align*}
- \begin{align*}y=3.75\end{align*}

Identify the slope of the following lines.

Identify the slope and \begin{align*}y-\end{align*}intercept for the following functions.

For 7-10, plot the following functions on a graph.

- \begin{align*}y=2x+5\end{align*}
- \begin{align*}y=-0.2x+7\end{align*}
- \begin{align*}y=x\end{align*}
- \begin{align*}y=3.75\end{align*}

- Which two of the following lines are parallel?
- \begin{align*}y=2x+5\end{align*}
- \begin{align*}y=-0.2x+7\end{align*}
- \begin{align*}y=x\end{align*}
- \begin{align*}y=3.75\end{align*}
- \begin{align*}y= -\frac{1}{5}x-11\end{align*}
- \begin{align*}y=-5x+5\end{align*}
- \begin{align*}y=-3x+11\end{align*}
- \begin{align*}y=3x+3.5\end{align*}

- What is the \begin{align*}y-\end{align*}intercept of the line passing through (1, -4) and (3, 2)?
- What is the \begin{align*}y-\end{align*}intercept of the line with slope -2 that passes through (3, 1)?
- Line \begin{align*}A\end{align*} passes through the points (2, 6) and (-4, 3). Line \begin{align*}B\end{align*} passes through the point (3, 2.5), and is parallel to line \begin{align*}A\end{align*}
- Write an equation for line \begin{align*}A\end{align*} in slope-intercept form.
- Write an equation for line \begin{align*}B\end{align*} in slope-intercept form.

- Line \begin{align*}C\end{align*} passes through the points (2, 5) and (1, 3.5). Line \begin{align*}D\end{align*} is parallel to line \begin{align*}C\end{align*}, and passes through the point (2, 6). Name another point on line \begin{align*}D\end{align*}. (Hint: you can do this without graphing or finding an equation for either line.)