ORCCA Slope-Intercept Form

Section 3.5 Slope-Intercept Form

Objectives: PCC Course Content and Outcome Guide

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 $\$50\text{,}$ she has

\begin{equation*} y=20x+50 \end{equation*}

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 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 illustrates this in the abstract.

Figure 3.5.2. Yara’s savings

a coordinate plane with the line y=20x+50; the line extends to the left and the right — Figure 3.5.3. Generic line with slope and $y$-intercept

We already have an accepted symbol, $m\text{,}$ for the slope of a line. The $y$-intercept 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 $0\text{.}$ 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.

What was the value of $b$ in the plot of Yara’s savings?

What is the $y$-intercept?

Explanation.

The line crosses the $y$-axis at $(0,50)\text{,}$ so the value of $b$ is $50\text{.}$ And the $y$-intercept is $(0,50)\text{.}$

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

\begin{equation*} y=20x+50 \end{equation*}

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

\begin{equation} y=mx+b\tag{3.5.1} \end{equation}

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

Checkpoint 3.5.6.

What are the slope and $y$-intercept for each of the following line equations?

Equation	Slope	$y$-intercept
$y={3.1x+1.78}$
$y={-17x+112}$
$y={\frac{3}{7}x-\frac{2}{3}}$
$y={13-8x}$
$y={1-\frac{2x}{3}}$
$y={2x}$
$y={3}$

Explanation.

In the first three equations, simply read the slope $m$ according to slope-intercept form. The slopes are $3.1\text{,}$ $-17\text{,}$ and ${\frac{3}{7}}\text{.}$

The fourth equation was written with the terms not in the slope-intercept form order. It could be written $y=-8x+13\text{,}$ and then it is clear that its slope is $-8\text{.}$ In any case, the slope is the coefficient of $x\text{.}$

The fifth equation is also written with the terms not in the slope-intercept form order. Changing the order of the terms, it could be written $y=-\frac{2x}{3}+1\text{,}$ but this still does not match the pattern of slope-intercept form. Considering how fraction multiplication works, $\frac{2x}{3}=\frac{2}{3}\cdot\frac{x}{1}=\frac{2}{3}x\text{.}$ So we can write this equation as $y=-\frac{2}{3}x+1\text{,}$ and we see the slope is $-\frac{2}{3}\text{.}$

The last two equations could be written $y=2x+0$ and $y=0x+3\text{,}$ allowing us to read their slopes as $2$ and $0\text{.}$

For the $y$-intercepts, remember that we are expected to answer using an ordered pair $(0,b)\text{,}$ not just a single number $b\text{.}$ We can simply read that the first two $y$-intercepts are ${\left(0,1.78\right)}$ and ${\left(0,112\right)}\text{.}$

The third equation does not exactly match the slope-intercept form, until you view it as $y=\frac{3}{7}x+\left(-\frac{2}{3}\right)\text{,}$ and then you can see that its $y$-intercept is $\left(0,-\frac{2}{3}\right)\text{.}$

With the fourth equation, after rewriting it as $y=-8x+13\text{,}$ we can see that its $y$-intercept is $(0,13)\text{.}$

We already explored rewriting the fifth equation as $y=-\frac{2}{3}x+1\text{,}$ where we can see that its $y$-intercept is $(0,1)\text{.}$

The last two equations could be written $y=2x+0$ and $y=0x+3\text{,}$ allowing us to read their $y$-intercepts as $(0,0)$ and $(0,3)\text{.}$

Alternatively, we know that $y$-intercepts happen where $x=0\text{,}$ and substituting $x=0$ into each equation gives you the $y$-value of the $y$-intercept.

Remark 3.5.7.

The number $b$ is the $y$-value when $x=0\text{.}$ Therefore it is common to refer to $b$ as the initial value or starting value of a linear relationship.

Example 3.5.8.

With a simple equation like $y=2x+3\text{,}$ we can see that this is a line whose slope is $2$ and which has initial value $3\text{.}$ So starting at $y=3$ on the $y$-axis, each time we increase the $x$-value by $1\text{,}$ the $y$-value increases by $2\text{.}$ 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
increase by $1\longrightarrow$	$1$	$5$	increase $\longleftarrow$ by $2$
increase by $1\longrightarrow$	$2$	$7$	increase $\longleftarrow$ by $2$
increase by $1\longrightarrow$	$3$	$9$	increase $\longleftarrow$ by $2$
increase by $1\longrightarrow$	$4$	$11$	increase $\longleftarrow$ by $2$

$A Cartesian graph with the points plotted from the table;there is a line passing through the points and slope triangles drawn between each pair of points$

Example 3.5.9.

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

$x$-values	$y$-values
$0$	$-4$
$2$	$2$
$5$	$11$
$9$	$23$

Explanation.

To assess whether the relationship is linear, we have to recall from Section 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 $2\text{,}$ $y$ increases by $6\text{.}$ The first rate of change is $\frac{6}{2}=3\text{.}$
When $x$ increases by $3\text{,}$ $y$ increases by $9\text{.}$ The second rate of change is $\frac{9}{3}=3\text{.}$
When $x$ increases by $4\text{,}$ $y$ increases by $12\text{.}$ The third rate of change is $\frac{12}{4}=3\text{.}$

Since the rates of change are all the same, $3\text{,}$ the relationship is linear and the slope $m$ is $3\text{.}$ According to the table, when $x=0\text{,}$ $y=-4\text{.}$ So the starting value, $b\text{,}$ is $-4\text{.}$ So in slope-intercept form, the line’s equation is $y=3x-4\text{.}$

Checkpoint 3.5.10.

Decide whether data in the table has a linear relationship. If so, write the linear equation in slope-intercept form. This may not be as easy as the previous example. Read the solution for a full explanation.

$x$-values	$y$-values
$3$	$-2$
$6$	$-8$
$8$	$-12$
$11$	$-18$

The data

does
does not

have a linear relationship, because:

changes in x are not constant
rates of change between data points are constant
rates of change between data points are not constant

The slope-intercept form of the equation for this line is .

Explanation.

To assess whether the relationship is linear, we examine rates of change between data points.

The first rate of change is $\frac{-6}{3}=-2\text{.}$
The second rate of change is $\frac{-4}{2}=-2\text{.}$
The third rate of change is $\frac{-6}{3}=-2\text{.}$

Since the rates of change are all the same, $-2\text{,}$ the relationship is linear and the slope $m$ is $-2\text{.}$

So we know that the slope-intercept equation is $y=-2x+b\text{,}$ but what number is $b\text{?}$ The table does not directly tell us what the initial $y$-value is. One approach is to use any point that we know the line passes through, and use algebra to solve for $b\text{.}$ We know the line passes through $(3,-2)\text{,}$ so

\begin{equation*} \begin{aligned} y\amp=-2x+b\\ \substitute{-2}\amp=-2(\substitute{3})+b\\ -2\amp=-6+b\\ 4\amp=b \end{aligned} \end{equation*}

So the equation is $y=-2x+4\text{.}$

Subsection 3.5.2 Graphing Slope-Intercept Equations

Example 3.5.11.

The conversion formula for a Celsius temperature into Fahrenheit is $F=\frac{9}{5}C+32\text{.}$ This appears to be in slope-intercept form, except that $x$ and $y$ are replaced with $C$ and $F\text{.}$ Suppose you are asked to graph this equation. How will you proceed? You could make a table of values as we do in Section 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 $(0,32)\text{.}$

Set up the axes using an appropriate window and labels. Considering the freezing temperature of water ($0^{\circ}$ Celsius or $32^{\circ}$ Fahrenheit), and the boiling temperature of water ($100^{\circ}$ Celsius or $212^{\circ}$ Fahrenheit), it’s reasonable to let $C$ run through at least $0$ to $100$ and $F$ run through at least $32$ to $212\text{.}$
Plot the $F$-intercept, which is at $(0,32)\text{.}$
Starting at the $F$-intercept, use slope triangles to reach the next point. Since our slope is $\frac{9}{5}\text{,}$ that suggests a “run” of $5$ and a “rise” of $9$ might work. But as Figure 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 $\frac{18}{10}\text{,}$ $\frac{90}{50}\text{,}$ $\frac{900}{50}\text{,}$ or $\frac{45}{25}$ which all reduce to $\frac{9}{5}\text{.}$ 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 $90\text{.}$
Connect your points with a straight line, use arrowheads, and label the equation.

a coordinate plane with the first quadrant shown; the scale on the x-axis is 25 degrees and the scale on the y-axis is 50 degrees; the y-intercept of (0,32) is graphed — Figure 3.5.12. Graphing $F=\frac{9}{5}C+32$

the previous graph with small slope triangles shown with a run of 5 and a rise of 9; the larger slope triangles with a run of 50 and a rise of 90 get us to the next points which are (50,122) and (100,212) — Figure 3.5.12. Graphing $F=\frac{9}{5}C+32$

Example 3.5.13.

Graph $y=-\frac{2}{3}x+10\text{.}$

a coordinate plane with a scale of 2 on the x-axis and 5 on the y-axis; the y-intercept of (0,10) is plotted — (a) Setting up the axes in an appropriate window and making sure that the $y$-intercept will be visible, and that any “run” and “rise” amounts we wish to use will not make triangles that are too big or too small.

the previous graph with slope triangles drawn; to the right the run is 3 and the rise is -2; to the left the run is -3 and the rise is 2 — (a) Setting up the axes in an appropriate window and making sure that the $y$-intercept will be visible, and that any “run” and “rise” amounts we wish to use will not make triangles that are too big or too small.

Example 3.5.15.

Graph $y=3x+5\text{.}$

a coordinate plane with a scale of 2 on the x-axis and 5 on the y-axis; the y-intercept of (0,5) is plotted — (a) Setting up the axes to make sure that the $y$-intercept will be visible, and that any “run” and “rise” amounts we wish to use will not make triangles that are too big or too small.

the previous graph with slope triangles drawn; to the right the run is 1 and the rise is 3; to the left the run is -1 and the rise is -3 — (a) Setting up the axes to make sure that the $y$-intercept will be visible, and that any “run” and “rise” amounts we wish to use will not make triangles that are too big or too small.

Subsection 3.5.3 Writing a Slope-Intercept Equation Given a Graph

We can write a linear equation in slope-intercept form based on its graph. We need to be able to calculate the line’s slope and see its $y$-intercept.

Checkpoint 3.5.17.

Use the graph to write an equation of the line in slope-intercept form.

Explanation.

On the line, pick two points with easy-to-read integer coordinates so that we can calculate slope. It doesn’t matter which two points we use; the slope will be the same.

Using the slope triangle, we can calculate the line’s slope:

\begin{equation*} \text{slope}=\frac{\Delta y}{\Delta x}=\frac{-2}{4}=-\frac{1}{2} \text{.} \end{equation*}

From the graph, we can see the $y$-intercept is $(0,6)\text{.}$

With the slope and $y$-intercept found, we can write the line’s equation:

\begin{equation*} y=-\frac{1}{2}x+6 \text{.} \end{equation*}

Checkpoint 3.5.18.

There are seven public four-year colleges in Oregon. The graph plots the annual in-state tuition for each school on the $x$-axis, and the median income of former students ten years after first enrolling on the $y$-axis.

Write an equation for this line in slope-intercept form.

Explanation.

Do your best to identify two points on the line. We go with $(0,27500)$ and $(11000,45000)\text{.}$

\begin{equation*} \frac{\Delta y}{\Delta x}=\frac{45000-27500}{11000-0}=\frac{17500}{11000}\approx1.591 \end{equation*}

So the slope is about $1.591$ dollars of median income per dollar of tuition. This is only an estimate since we are not certain the two points we chose are actually on the line.

Estimating the $y$-intercept to be at $(0,27500)\text{,}$ we have $y=1.591x+27500\text{.}$

Subsection 3.5.4 Writing a Slope-Intercept Equation Given Two Points

The idea that any two points uniquely determine a line has been understood for thousands of years in many cultures around the world. Once you have two specific points, there is a straightforward process to find the slope-intercept form of the equation of the line that connects them.

Example 3.5.19.

Find the slope-intercept form of the equation of the line that passes through the points $(0,5)$ and $(8,-5)\text{.}$

Explanation.

We are trying to write down $y=mx+b\text{,}$ but with specific numbers for $m$ and $b\text{.}$ So the first step is to find the slope, $m\text{.}$ To do this, recall the slope formula from Section 4. It says that if a line passes through the points $(x_1,y_1)$ and $(x_2,y_2)\text{,}$ then the slope is found by the formula $m=\frac{y_2-y_1}{x_2-x_1}\text{.}$

Applying this to our two points $(\overset{x_1}{0},\overset{y_1}{5})$ and $(\overset{x_2}{8},\overset{y_2}{-5})\text{,}$ we see that the slope is:

\begin{align*} m\amp=\frac{y_2-y_1}{x_2-x_1}\\ \amp=\frac{\substitute{-5}-\substitute{5}}{\substitute{8}-\substitute{0}}\\ \amp=\frac{-10}{8}=-\frac{5}{4} \end{align*}

We are trying to write $y=mx+b\text{.}$ Since we already found the slope, we know that we want to write $y=-\frac{5}{4}x+b$ but we need a specific number for $b\text{.}$ We happen to know that one point on this line is $(0,5)\text{,}$ which is on the $y$-axis because its $x$-value is $0\text{.}$ So $(0,5)$ is this line’s $y$-intercept, and therefore $b=5\text{.}$ So, our equation is

\begin{equation*} y=-\frac{5}{4}x+5\text{.} \end{equation*}

Example 3.5.20.

Find the slope-intercept form of the equation of the line that passes through the points $(3,-8)$ and $(-6,1)\text{.}$

Explanation.

The first step is always to find the slope between our two points: $(\overset{x_1}{3},\overset{y_1}{-8})$ and $(\overset{x_2}{-6},\overset{y_2}{1})\text{.}$ Using the slope formula again, we have:

\begin{align*} m\amp=\frac{y_2-y_1}{x_2-x_1}\\ \amp=\frac{\substitute{1}-\substitute{(-8)}}{\substitute{{-}6}-\substitute{3}}\\ \amp=\frac{9}{-9}\\ \amp=-1 \end{align*}

Now that we have the slope, we can write $y=-1x+b\text{,}$ which simplifies to $y=-x+b\text{.}$ Unlike in Example 19, we are not given the value of $b$ because neither of our two given points have an $x$-value of $0\text{.}$ The trick to finding $b$ is to remember that we have two points that we know make the equation true! This means all we have to do is substitute either point into the equation for $x$ and $y$ and solve for $b\text{.}$ Let’s arbitrarily choose $(3,-8)$ to plug in.

\begin{align*} y\amp=-x+b\\ \substitute{-8}\amp=-(\substitute{3})+b\amp\text{(Now solve for }b\text{.)}\\ -8\amp=-3+b\\ -8\addright{3}\amp=-3+b\addright{3}\\ -5\amp=b \end{align*}

In conclusion, the equation for which we were searching is $y=-x-5\text{.}$

Don’t be tempted to plug in values for $x$ and $y$ at this point. The general equation of a line in any form should have (at least one, and in this case two) variables in the final answer.

Checkpoint 3.5.21.

Find the slope-intercept form of the equation of the line that passes through the points $(-3,150)$ and $(0,30)\text{.}$

Explanation.

The first step is always to find the slope between our points: $(\overset{x_1}{-3},\overset{y_1}{150})$ and $(\overset{x_2}{0},\overset{y_2}{30})\text{.}$ Using the slope formula, we have:

\begin{equation*} \begin{aligned} m\amp=\frac{y_2-y_1}{x_2-x_1}\\ \amp=\frac{\substitute{30}-\substitute{150}}{\substitute{0}-\substitute{(-3)}}\\ \amp=\frac{-120}{3}\\ \amp=-40 \end{aligned} \end{equation*}

Now we can write $y=-40x+b$ and to find $b$ we need look no further than one of the given points: $(0,30)\text{.}$ Since the $x$-value is $0\text{,}$ the value of $b$ must be $30\text{.}$ So, the slope-intercept form of the line is

\begin{equation*} y=-40x+30 \end{equation*}

Checkpoint 3.5.22.

Find the slope-intercept form of the equation of the line that passes through the points $\left(-3,\frac{3}{4}\right)$ and $\left(-6,-\frac{17}{4}\right)\text{.}$

Explanation.

First find the slope through our points: $\left(-3,\frac{3}{4}\right)$ and $\left(-6,-\frac{17}{4}\right)\text{.}$

\begin{equation*} \begin{aligned} m\amp=\frac{y_2-y_1}{x_2-x_1}\\ \amp=\frac{\substitute{-\frac{17}{4}}-\substitute{\frac{3}{4}}}{\substitute{-6}-\substitute{(-3)}}\\ \amp=\frac{\frac{-20}{4}}{-3}\\ \amp=\frac{-5}{-3}\\ \amp=\frac{5}{3} \end{aligned} \end{equation*}

So far we have $y=\frac{5}{3}x+b\text{.}$ Now we need to solve for $b$ since neither of the points given were the vertical intercept. Recall that to do this, we will choose one of the two points and plug it into our equation. We choose $\left(-3,\frac{3}{4}\right)\text{.}$

\begin{equation*} \begin{aligned} y\amp=\frac{5}{3}x+b\\ \substitute{\frac{3}{4}}\amp=\frac{5}{3}(\substitute{-3})+b\\ \frac{3}{4}\amp=-5+b\\ \frac{3}{4}\addright{5}\amp=-5+b\addright{5}\\ \frac{3}{4}+\frac{20}{4}\amp=b\\ \frac{23}{4}\amp=b \end{aligned} \end{equation*}

Lastly, we write our equation.

\begin{equation*} y=\frac{5}{3}x+\frac{23}{4} \end{equation*}

Subsection 3.5.5 Modeling with Slope-Intercept Form

We can model many relatively simple relationships using slope-intercept form, and then solve related questions using algebra. Here are a few examples.

Example 3.5.23.

Uber is a ride-sharing company. Its pricing in Portland factors in how much time and how many miles a trip takes. But if you assume that rides average out at a speed of 30 mph, then their pricing scheme boils down to a base of $\$7.35$ for the trip, plus $\$3.85$ per mile. Use a slope-intercept equation and algebra to answer these questions.

How much is the fare if a trip is $5.3$ miles long?
With $\$100$ available to you, how long of a trip can you afford?

Explanation.

The rate of change (slope) is $\$3.85$ per mile, and the starting value is $\$7.35\text{.}$ So the slope-intercept equation is

\begin{equation*} y=3.85x+7.35\text{.} \end{equation*}

In this equation, $x$ stands for the number of miles in a trip, and $y$ stands for the amount of money to be charged.

If a trip is $5.3$ miles long, we substitute $x=5.3$ into the equation and we have:

\begin{align*} y\amp=3.85x+7.35\\ \amp=3.85(\substitute{5.3})+7.35\\ \amp=20.405+7.35\\ \amp=27.755 \end{align*}

And the $5.3$-mile ride will cost you about $\$27.76\text{.}$ (We say “about,” because this was all assuming you average 30 mph.)

Next, to find how long of a trip would cost $\$100\text{,}$ we substitute $y=100$ into the equation and solve for $x\text{:}$

\begin{align*} y\amp=3.85x+7.35\\ \substitute{100}\amp=3.85x+7.35\\ 100\subtractright{7.35}\amp=3.85x\\ 92.65\amp=3.85x\\ \divideunder{92.65}{3.85}\amp=x\\ 24.06\amp\approx x \end{align*}

So with $\$100$ you could afford a little more than a $24$-mile trip.

Checkpoint 3.5.24.

In a certain wildlife reservation in Africa, there are approximately $2400$ elephants. Sadly, the population has been decreasing by $30$ elephants per year. Use a slope-intercept equation and algebra to answer these questions.

If the trend continues, what would the elephant population be $15$ years from now?
If the trend continues, how many years will it be until the elephant population dwindles to $1200\text{?}$

Explanation.

The rate of change (slope) is $-30$ elephants per year. Notice that since we are losing elephants, the slope is a negative number. The starting value is $2400$ elephants. So the slope-intercept equation is

\begin{equation*} y=-30x+2400 \text{.} \end{equation*}

In this equation, $x$ stands for a number of years into the future, and $y$ stands for the elephant population. To estimate the elephant population $15$ years later, we substitute $x$ in the equation with $15\text{,}$ and we have:

\begin{equation*} \begin{aligned} y\amp=-30x+2400\\ \amp=-30(\substitute{15})+2400\\ \amp=-450+2400\\ \amp=1950 \end{aligned} \end{equation*}

So if the trend continues, there would be $1950$ elephants on this reservation 15 years later.

Next, to find when the elephant population would decrease to $1200\text{,}$ we substitute $y$ in the equation with $1200\text{,}$ and solve for $x\text{:}$

\begin{equation*} \begin{aligned} y\amp=-30x+2400\\ \substitute{1200}\amp=-30x+2400\\ 1200\subtractright{2400}\amp=-30x\\ -1200\amp=-30x\\ \divideunder{-1200}{-30}\amp=x\\ 40\amp=x \end{aligned} \end{equation*}

So if the trend continues, $40$ years later, the elephant population would dwindle to $1200\text{.}$

Reading Questions 3.5.6 Reading Questions

1.

How does “slope-intercept form” get its name?

2.

What are two phrases you can use for “$b$” in a slope-intercept form line equation?

3.

Explain the two basic steps to graphing a line when you have the equation in slope-intercept form. (Not counting the step where you draw and label the axes and ticks.)

Exercises 3.5.7 Exercises

Review and Warmup

Exercise Group.

1.

Evaluate ${9c+8a}$ for $c = 4$ and $a = 2\text{.}$

2.

Evaluate ${-6c-9B}$ for $c = -8$ and $B = -6\text{.}$

3.

Evaluate

\begin{equation*} \displaystyle\frac{y_2 - y_1}{x_2 - x_1} \end{equation*}

for $x_1 = 3\text{,}$ $x_2 = -5\text{,}$ $y_1 = 1\text{,}$ and $y_2 = 17\text{:}$

Answer:

4.

Evaluate

\begin{equation*} \displaystyle\frac{y_2 - y_1}{x_2 - x_1} \end{equation*}

for $x_1 = 8\text{,}$ $x_2 = -19\text{,}$ $y_1 = 18\text{,}$ and $y_2 = 6\text{:}$

Answer:

Skills Practice

This line’s $y$-intercept is .

Graphs and Slope-Intercept Form.

21.

A line’s graph is given. What is this line’s slope-intercept equation?

22.

A line’s graph is given. What is this line’s slope-intercept equation?

23.

A line’s graph is given. What is this line’s slope-intercept equation?

24.

A line’s graph is given. What is this line’s slope-intercept equation?

25.

A line’s graph is given. What is this line’s slope-intercept equation?

26.

A line’s graph is given. What is this line’s slope-intercept equation?

27.

A line’s graph is given. What is this line’s slope-intercept equation?

28.

A line’s graph is given. What is this line’s slope-intercept equation?

Graph Equations.

Graph the equation.

29.

${y = 3x}$

30.

$y=5x$

31.

$y=-3x$

32.

$y=-2x$

33.

$y=\frac{5}{2}x$

34.

$y=\frac{1}{4}x$

35.

$y=-\frac{1}{3}x$

36.

$y=-\frac{5}{4}x$

37.

$y=5x+2$

38.

$y=3x+6$

39.

$y=-4x+3$

40.

$y=-2x+5$

41.

$y=x-4$

42.

$y=x+2$

43.

$y=-x+3$

44.

$y=-x-5$

45.

$y=\frac{2}{3}x+4$

46.

$y=\frac{3}{2}x-5$

47.

$y=-\frac{3}{5}x-1$