# 7.6: Graphing Equations in Slope-Intercept Form - Mathematics

## Using the Slope and Intercept to Graph a Line

When a linear equation is given in the general form, (ax+by=c), we observed that an efficient graphical approach was the intercept method. We let (x=0) and computed the corresponding value of (y), then let (y=0) and computed the corresponding value of (x).

When an equation is written in the slope-intercept form, (y=mx+b), there are also efficient ways of constructing the graph. One way, but less efficient, is to choose two or three (x)-values and compute to find the corresponding (y)-values. However, computations are tedious, time consuming, and can lead to errors. Another way, the method listed below, makes use of the slope and the (y)-intercept for graphing the line. It is quick, simple, and involves no computations.

Graphing Method

1. Plot the (y)-intercept ((0, b)).
2. Determine another point by using the slope m.
3. Draw a line through the two points.

Recall that we defined the slope (m) as the ratio (dfrac{y_2-y_1}{x_2-x_1}). The numerator (y_2−y_1) represents the number of units that (y) changes and the denominator (x_2 - x_1) represents the number of units that (x) changes. Suppose (m=pq). Then (p) is the number of units that (y) changes and (q) is the number of units that (x) changes. Since these changes occur simultaneously, start with your pencil at the (y)-intercept, move (p) units in the appropriate vertical direction, and then move (q) units in the appropriate horizontal direction. Mark a point at this location.

## Sample Set A

Graph the following lines.

Example (PageIndex{1})

(y = dfrac{3}{4}x + 2)

1. The (y)-intercept is the point ((0,2)). Thus the line crosses the (y)-axis (2) units above the origin. Mark a point at ((0,2)).

2. The slope, (m), is (dfrac{3}{4}). This means that if we start at any point on the line and move our pencil (3) units up and then (4) units to the right, we’ll be back on the line. Start at a known point, the (y)-intercept ((0, 2)). Move up (3) units, then move (4) units to the right. Mark a point at this location. (Note also that dfrac{3}{4} = dfrac{-3}{-4}). This means that if we start at any point on the line and move our pencil (3) units down and (4) units to the left, we’ll be back on the line. Note also that (dfrac{3}{4} = dfrac{dfrac{3}{4}}{1}). This means that if we start at any point on the line and move to the right (1) unit, we’ll have to move up (dfrac{3}{4}) unit to get back on the line.)

3. Draw a line through both points.

Example (PageIndex{2})

(y = -dfrac{1}{2}x + dfrac{7}{2})

1. The (y)-intercept is the point ((0, dfrac{7}{2})). Thus the line crosses the (y)-axis (dfrac{7}{2}) units above the origin. Mark a point ((0, dfrac{7}{2})), or ((0, 3dfrac{1}{2})).

2. The slope, (m), is (-dfrac{1}{2}). We can write (-dfrac{1}{2}) as (dfrac{-1}{2}). Thus, we start at a known point, the (y)-intercept ((0, 3dfrac{1}{2})), move down one unit (because of the (-1)), then move right (2) units. Mark a point at this location.

3. Draw a line through both points.

Example (PageIndex{3})

(y = dfrac{2}{5}x)

1. We can put this equation into explicit slope-intercept by writing it as (y = dfrac{2}{5}x + 0).

The (y)-intercept is at the point ((0, 0)), the origin. This line goes right through the origin.

2. The slope, (m), is (dfrac{2}{5}). Starting at the origin, we move up (2) units, then move to the right (5) units. Mark a point at this location.

3. Draw a line through the two points.

Example (PageIndex{4})

(y = 2x - 4)

1. The (y)-intercept is the point ((0, -4)). Thus the line crosses the (y)-axis (4) units below the origin. Mark a point at ((0, -4)).

2. The slope, (m), is (2). If we write the slope as a fraction, (2 = dfrac{2}{1}), we can read how to make the changes. Start at the known point ((0, -4)), move up (2) units, then move right (1) unit. Mark a point at this location.

3. Draw a line through the two points.

## Practice Set A

Use the (y)-intercept and the slope to graph each line.

Practice Problem (PageIndex{1})

(y = dfrac{-2}{3} + 4)

Practice Problem (PageIndex{2})

(y = dfrac{3}{4}x)

## Exercises

For the following problems, graph the equations.

Exercise (PageIndex{1})

(y = dfrac{2}{3} + 1)

Exercise (PageIndex{2})

(y = dfrac{1}{4}x - 2)

Exercise (PageIndex{3})

(y = 5x - 4)

Exercise (PageIndex{4})

(y = -dfrac{6}{5} - 3)

Exercise (PageIndex{5})

(y = dfrac{3}{2} - 5)

Exercise (PageIndex{6})

(y = dfrac{1}{5}x + 2)

Exercise (PageIndex{7})

(y = -dfrac{8}{3} + 4)

Exercise (PageIndex{8})

(y = -dfrac{10}{3} + 6)

Exercise (PageIndex{9})

(y = 1x - 4)

Exercise (PageIndex{10})

(y = -2x + 1)

Exercise (PageIndex{11})

(y = x + 2)

Exercise (PageIndex{12})

(y = dfrac{3}{5}x)

Exercise (PageIndex{13})

(y = -dfrac{4}{3})

Exercise (PageIndex{14})

(y = x)

Exercise (PageIndex{15})

(y = -x)

Exercise (PageIndex{16})

(3y−2x=−3)

Exercise (PageIndex{17})

(6x+10y=30)

Exercise (PageIndex{18})

(x+y=0)

## Exercises for Review

Exercise (PageIndex{19})

Solve the inequality (2 - 4x ge x - 3)

(x≤1)

Exercise (PageIndex{20})

Graph the inequality (y+3>1.)

Exercise (PageIndex{21})

Graph the equation (y = -2).

Exercise (PageIndex{22})

Determine the slope and (y)-intercept of the line (−4y−3x=16).

Exercise (PageIndex{23})

Find the slope of the line passing through the points ((−1, 5)) and ((2, 3)).

(m = dfrac{-2}{3})

## SOLUTION: Write an equation for the line described as slope=-3/5 and passes through (7,6). Write in slope intercept form.

You can put this solution on YOUR website!
Write an equation for the line described as slope=-3/5 and passes through (7,6). Write in slope intercept form.

Step 1. The slope-intercept form is given as y=mx+b where m is the slope and b is the y-intercept b at x=0 or point (0,b). Here, the slope m=-3/5.

Step 2. The slope m is given as

Step 3. Let (x1,y1)=(6,7) or x1=7 and y1=6. Let other point be (x2,y2)=(x,y) or x2=x and y2=y.

Step 4. Now we're given . Substituting above values and variables in the slope equation m yields the following steps:

Step 5. Multiply (x-7) to both sides to get rid of denominators on both sides of equation.

Step 6. Now add +6 to both sides of equation to solve for y.

Step 7. ANSWER: The equation is

Here's a graph below and note the slope and y-intercept at x=0 or point (0, 51/5) and the x-intercept at y=0 or at point (17, 0)and note it is consistent with the equation when substituting these

This slope-intercept game has ten multiple choice problems about the slope-intercept form of a linear equation.
Here are some important facts about linear equations that you should know:

The slope-intercept formula of a linear equation is y= mx + b (where m represents the slope and b represents the y-intercept).
The slope is the rise (the vertical change) over the run (the horizontal change).
The y-intercept of a line is the y-coordinate of the point of intersection between the graph of the line and the y-intercept.

You can play this game alone, with a friend, or in two teams. This game is a multi-player game that can be played on computers, Promethean boards, smart boards, iPads, and other tablets. You do not need to install an app to play this game on the iPad. Have fun evaluating algebraic expressions!

The game is based on the following Common Core Math Standards:

CCSS8.F.3
Interpret the equation y = mx + b as defining a linear function, whose
graph is a straight line give examples of functions that are not linear.,

CCSS8.F.4
Construct a function to model a linear relationship between two
quantities. Determine the rate of change and initial value of the
function from a description of a relationship or from two (x, y) values,
including reading these from a table or from a graph. Interpret the rate
of change and initial value of a linear function in terms of the situation
it models, and in terms of its graph or a table of values.

## 7.6: Graphing Equations in Slope-Intercept Form - Mathematics

To graph a straight line we need at least two points which lie on the straight line. From the slope-intercept form of the given straight line, we can calculate two points on the line very easily using the information present in the equation. Consider a straight line with slope m and y-intercept c. We know the slope-intercept form of the equation of this straight line is: y = mx + c. Here y-intercept c indicates that the line has cut the Y-axis on a distance of c from the origin, which further implies that the cutting point is (0, c). Now we have a fixed point on the line which we can plot in the graph. Using this point and the slope of the straight line we can draw a graph of the straight line.

### Graphing slope-intercept form

Now we have a fixed point on the line which is the cutting point with coordinate (0, c). To determine the second point we need to use the slope of the line which is m for the considered straight line.

If m is in fraction form then p is the numerator and q is the denominator. If m is an integer then we can always take m as numerator(p) and 1 as denominator(q)(as m/1 = m).

which further implies that q units change in x−coordinate will result p units change in y−coordinate. Thus, if point (x, y) lies on the straight line, then point (x + q, y + p) will also lie on that line. For the considered line, we know (0, c) lies on the line and p/q (m = p/q) is the slope of the line. As discussed above, point (0 + q, c + p) or point (q, c + p) will also lie on that line. Now we can plot these two points on the graph and joining them will give us the required straight line. For further clarification, see above image where m = p/q = 3/2 and c = 4.

### Sample Problems on Graph Plotting

Problem 1: Plot a Graph for Slope intercept equation, y = 3x – 5.

Comparing given equation with y = mx + c we get,

m = 3 ⇒ p/q = 3/1 ⇒ p = 3, q = 1 and c = -5

c = -5 implies that y-intercept of the line is -5, i.e.

the line cuts the Y-axis on the point (0, -5)

m = 3 or p/q = 3/1 implies that slope of the line is 3,

i.e. 1 unit change in x-coordinate will result 3 units

change in y-coordinate. Thus, if point (x, y) lies on

the given straight line then point (x +1, y +3) will also

lie on that line.

⇒ Point (0, -5) and point (0+1, -5+3) ⇒(1, -2)

lies on the given straight line. Joining these two

points will give us the required line.

Problem 2: Plot a Graph for Slope intercept equation y = -(4/7)x + 2.

Comparing given equation with y = mx + c we get,

m = -(4/7) ⇒ p/q ⇒ -(4/7) ⇒ p = -4, q = 7 and c = 2

⇒ Point (0, 2) and point (0 + 7, 2 – 4) ⇒(7, -2)

lies on the given straight line. Joining these two

points will give us the required line.

Problem 3: Plot a Graph for Slope intercept equation y = (5/3)x + 4.

Comparing given equation with y = mx + c we get,

m = 5/3 ⇒ p/q ⇒ 5/3 ⇒ p = 5, q = 3 and c = 4

⇒ Point (0, 4) and point (0+3, 4+5) ⇒(3, 9)

lies on the given straight line. Joining these two

points will give us the required line.

## GRAPHING LINEAR EQUATIONS IN SLOPE INTERCEPT FORM WORKSHEET

Because slope '2' is a positive value, the line will be a rising line.

Because the y-intercept is 1, the line will intersect y-axis at 1.

Plot the y-intercept at (0, 1).

Because the run is 1, move 1 unit to the right from (0, 1).

Because the rise is 2 and the line is rising line, move 2 units up from the position reached in step 2.

Connect the points (0, 1) and (1, 3) to get the line.

Graph the following linear equation.

The equation 'y  =  -1.5x + 2' is in the form of

Because slope '-1.5' is a negative value, the line will be a falling line.

Because the y-intercept is 2, the line will intersect y-axis at 2.

Plot the y-intercept at (0, 2).

Because the run is 2,  move 2 units to the right from (0, 2).

Because the rise is 3 and the line is falling line, move 3 units down from the position reached in step 2.

Connect the points (0, 2) and (2, -1) to get the line.

Graph the following linear equation.

The equation 'x - y - 2  =  0' is not slope -intercept form.

Write the given equation in slope-intercept form.

The equation 'y  =  x - 2' is in the form of

Because slope '1' is a positive value, the line will be a rising line.

Because the y-intercept is -2, the line will intersect y-axis at -2.

Plot the y-intercept at (0, -2).

Because the run is 1,  move 1 unit to the right from (0, -2).

Because the rise is 1 and the line is rising line, move 1 unit up from the position reached in step 2.

Connect the points (0, -2) and (1, -1) to get the line.

Graph the following linear equation.

The equation '5x + y  =  3' is not slope -intercept form.

Write the given equation in slope-intercept form.

Subtract 5x from each side.

The equation 'y  =  -5x + 3' is in the form of

Because slope '-5' is a negative value, the line will be a falling line.

Because the y-intercept is 3, the line will intersect y-axis at 3.

Plot the y-intercept at (0, 3).

Because the run is 1,  move 1 unit to the right from (0, 3).

Because the rise is 5 and the line is falling line, move 5 units down from the position reached in step 2.

Connect the points (0, 3) and (1, -2) to get the line.

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## Finding the x Value Using Slope-Intercept Form

Although we mentioned that you wouldn't normally be the one to decide the value of y, there are cases where you will need to. Normally, you use this method when you want to find the x value given a particular y value. Using the slope-intercept form, this is a fairly easy thing to do if given the function and y-value. For instance, if you are presented with the function y = 6x - 1 and are told to find x when y is 11, you would plug in y, giving you 11 = 6x - 1. Then after adding 1 to each side of the equation and dividing by 6, you would get x = 2 .

## Open Resources for Community College Algebra

In this section, we will explore what is perhaps the most common way to write the equation of a line. It's known as slope-intercept form.

Figure 3.5.1. Alternative Video Lesson

### Subsection 3.5.1 Slope-Intercept Definition

Recall Example 3.4.5, where Yara had ($50) in her savings account when the year began, and decided to deposit ($20) each week without withdrawing any money. In that example, we model using (x) to represent how many weeks have passed. After (x) weeks, Yara has added (20x) dollars. And since she started with (\$50 ext<,>) she has

in her account after (x) weeks. In this example, there is a constant rate of change of (20) dollars per week, so we call that the slope as discussed in Section 3.4. We also saw in Figure 3.4.7 that plotting Yara's balance over time gives us a straight-line graph.

The graph of Yara's savings has some things in common with almost every straight-line graph. There is a slope, and there is a place where the line crosses the (y)-axis. Figure 3.5.3 illustrates this in the abstract.

We already have an accepted symbol, (m ext<,>) for the slope of a line. The is a point on the (y)-axis where the line crosses. Since it's on the (y)-axis, the (x)-coordinate of this point is (0 ext<.>) It is standard to call the point ((0,b)) the (y)-intercept, and call the number (b) the (y)-coordinate of the (y)-intercept.

###### Checkpoint 3.5.4 .

Use Figure 3.4.7 to answer this question.

One way to write the equation for Yara's savings was

where both (m=20) and (b=50) are immediately visible in the equation. Now we are ready to generalize this.

###### Definition 3.5.5 . Slope-Intercept Form.

When (x) and (y) have a linear relationship where (m) is the slope and ((0,b)) is the (y)-intercept, one equation for this relationship is

and this equation is called the of the line. It is called this because the slope and (y)-intercept are immediately discernible from the numbers in the equation.

###### Remark 3.5.7 .

The number (b) is the (y)-value when (x=0 ext<.>) Therefore it is common to refer to (b) as the or of a linear relationship.

###### Example 3.5.8 .

With a simple equation like (y=2x+3 ext<,>) we can see that this is a line whose slope is (2) and which has initial value (3 ext<.>) So starting at (y=3) on the (y)-axis, each time we increase the (x)-value by (1 ext<,>) the (y)-value increases by (2 ext<.>) With these basic observations, we can quickly produce a table and/or a graph.

 (x) (y) start on(y)-axis (longrightarrow) (0) (3) initial(longleftarrow) value increaseby (1longrightarrow) (1) (5) increase(longleftarrow) by (2) increaseby (1longrightarrow) (2) (7) increase(longleftarrow) by (2) increaseby (1longrightarrow) (3) (9) increase(longleftarrow) by (2) increaseby (1longrightarrow) (4) (11) increase(longleftarrow) by (2)
###### Example 3.5.9 .

Decide whether data in the table has a linear relationship. If so, write the linear equation in slope-intercept form (3.5.1).

To assess whether the relationship is linear, we have to recall from Section 3.3 that we should examine rates of change between data points. Note that the changes in (y)-values are not consistent. However, the rates of change are calculated as follows:

When (x) increases by (2 ext<,>) (y) increases by (6 ext<.>) The first rate of change is (frac<6><2>=3 ext<.>)

When (x) increases by (3 ext<,>) (y) increases by (9 ext<.>) The second rate of change is (frac<9><3>=3 ext<.>)

When (x) increases by (4 ext<,>) (y) increases by (12 ext<.>) The third rate of change is (frac<12><4>=3 ext<.>)

Since the rates of change are all the same, (3 ext<,>) the relationship is linear and the slope (m) is (3 ext<.>) According to the table, when (x=0 ext<,>) (y=-4 ext<.>) So the starting value, (b ext<,>) is (-4 ext<.>) So in slope-intercept form, the line's equation is (y=3x-4 ext<.>)

### Subsection 3.5.2 Graphing Slope-Intercept Equations

###### Example 3.5.11 .

The conversion formula for a Celsius temperature into Fahrenheit is (F=frac<9><5>C+32 ext<.>) This appears to be in slope-intercept form, except that (x) and (y) are replaced with (C) and (F ext<.>) Suppose you are asked to graph this equation. How will you proceed? You could make a table of values as we do in Section 3.2 but that takes time and effort. Since the equation here is in slope-intercept form, there is a nicer way.

Since this equation is for starting with a Celsius temperature and obtaining a Fahrenheit temperature, it makes sense to let (C) be the horizontal axis variable and (F) be the vertical axis variable. Note the slope is (frac<9><5>) and the vertical intercept (here, the (F)-intercept) is ((0,32) ext<.>)

Set up the axes using an appropriate window and labels. Considering the freezing temperature of water ((0^) Celsius or (32^) Fahrenheit), and the boiling temperature of water ((100^) Celsius or (212^) Fahrenheit), it's reasonable to let (C) run through at least (0) to (100) and (F) run through at least (32) to (212 ext<.>)

Plot the (F)-intercept, which is at ((0,32) ext<.>)

Starting at the (F)-intercept, use slope triangles to reach the next point. Since our slope is (frac<9><5> ext<,>) that suggests a “run” of (5) and a “rise” of (9) might work. But as Figure 3.5.12 indicates, such slope triangles are too tiny. You can actually use any fraction equivalent to (frac<9><5>) to plot using the slope, as in (frac<18><10> ext<,>) (frac<90><50> ext<,>) (frac<900><50> ext<,>) or (frac<45><25>) which all reduce to (frac<9><5> ext<.>) Given the size of our graph, we will use (frac<90><50>) to plot points, where we will try a “run” of (50) and a “rise” of (90 ext<.>)

Connect your points with a straight line, use arrowheads, and label the equation.

## Slope Intercept Equation of a Line

The equation of a line with a slope of m and a y-intercept of (0, b) is y = mx + b.

To graph a line that is written in slope-intercept form:

1. Plot the y-intercept on the coordinate plane.
2. Use the slope to find another point on the line.
3. Join the plotted points with a straight line.

This line is given in slope-intercept form: y = mx + b.

The y-intercept of the line is –3. b = –3

The slope of the line is 4. m = 4

Therefore the graph of the equation is shown below.

Determine the equation of the graph that is shown.

The y-intercept of the line is 1. b = 1.

The slope of the line is –2/3. m = –2/3.

Substitute these values into the slope-intercept equation of a line: y = mx + b

The equation of the line is .

In the xy-plane above, point C has coordinates (6, 9). Which of the following is an equation of the line that contains points O and C?

The graph shown above gives the relation between x° F and y° C. Which equation is represented by the graph?

## Slope-Intercept Form

Trying to write an equation in slope-intercept form? Have two points on your line? You'll need to find your slope and y-intercept. Watch this tutorial and see what needs to be done to write an equation in slope-intercept form!

#### How Do You Graph a Line If You're Given the Slope and the Intercept?

Trying to graph a line from a given slope and y-intercept? Think you need to find an equation first? Think again! In this tutorial, see how to use that given slope and y-intercept to graph the line.

#### How Do You Write an Equation of a Line in Slope-Intercept Form If You Have the Slope and the Y-Intercept?

Want to write an equation in slope-intercept form? Already have the slope and y-intercept? Perfect! Just correctly plug those values into your equation and you're done! Learn how in this tutorial.

#### What's Slope-Intercept Form of a Linear Equation?

When you're learning about linear equations, you're bound to run into the point-slope form of a line. This form is quite useful in creating an equation of a line if you're given the slope and a point on the line. Watch this tutorial, and learn about the point-slope form of a line!

#### How Do You Write the Equation of a Line in Slope-Intercept Form If You Have a Graph?

Working with the graph of a line? Trying to find the equation for that graph? Just pick two points on the line and use them to find the equation. This tutorial shows you how to take two points on the graph of a line and use them to find the slope-intercept form of the line!

#### How Do You Write the Equation of a Line in Slope-Intercept Form If You Have a Table?

Looking at a table of values that represents a linear equation? Want to find that equation? Then check out this tutorial! You'll see how to use values from a table to find the slope-intercept form of the line described in the table.

## Slope Intercept Form

The slope intercept form is probably the most frequently used way to express equation of a line. To be able to use slope intercept form, all that you need to be able to do is 1) find the slope of a line and 2) find the y-intercept of a line.

### The Formula

In general, the slope intercept form assumes the formula: y = mx + b .

##### Examples
• y = 5x + 3 is an example of the Slope Intercept Form and represents the equation of a line with a slope of 5 and and a y-intercept of 3.
• y = &minus2x + 6 represents the equation of a line with a slope of &minus2 and and a y-intercept of 6.

### Vertical and Horizontal lines

#### Vertical Lines

Since a vertical line goes straight up and down, its slope is undefined. Also, the x value of every point on a vertical line is the same. Therefore, whatever the x value is, is also the value of 'b'.

For instance, the red line in the picture below is the graph of the x = 1 .

#### Horizontal Lines

The equation of a horizontal line is y = b where b is the y-intercept.

Since the slope of a horizontal line is 0, the general formula for the standard form equation, y = mx + b becomes y = 0x + b y = b . Also,since the line is horizontal, every point on that line has the exact same y value. This y-value is therefore also the y-intercept. For instance, the red line in the picture below is the graph of the horizontal line y = 1 . (In depth lesson on the equation of a horizontal line)