
## Section4.5Slope-Intercept Form

###### ObjectivesPCC Course Content and Outcome Guide

In this section, we will explore one of the “standard” ways to write the equation of a line. It's known as slope-intercept form.

### Subsection4.5.1Slope-Intercept Definition

Recall Example 4.4.5, where Matthew had $\50$ in his 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, Matthew has added $20x$ dollars. And since he started with $\50\text{,}$ he has

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

in his 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 4.4. We also saw in Figure 4.4.7 that plotting Matthew's balance over time gives us a straight-line graph.

The graph of Matthew'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 4.5.2 illustrates this in the abstract.

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 $y$-intercept $(0,b)$ where $b$ represents the position of the $y$-intercept on the $y$-axis.

###### Checkpoint4.5.3

Use Figure 4.4.7 to answer this question.

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

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

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

###### Definition4.5.4Slope-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

$$y=mx+b\label{equation-slope-intercept-form}\tag{4.5.1}$$

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.

###### Remark4.5.6

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.

###### Example4.5.7

Let's review. With a simple equation like $y=2x+3\text{,}$ we can see that there is a line whose slope is $2$ and which has initial value $3\text{.}$ So starting at $y=3$ when $x=0$ (that is, on the $y$-axis), each time you would increase the $x$-value by $1\text{,}$ the $y$-value increases by $2\text{.}$ With these basic observations, you may 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 $2$
###### Example4.5.8

Decide whether data in the table has a linear relationship. If so, write the linear equation in slope-intercept form (4.5.1).

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

To assess whether the relationship is linear, we have to recall from Section 4.3 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 thusly:

• 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 equation is $y=3x-4\text{.}$

### Subsection4.5.2Graphing Slope-Intercept Equations

###### Example4.5.11

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 4.2 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 $y$-intercept is $(0,32)\text{.}$

1. Set up the axes using an appropriate window and labels. Considering the freezing and boiling temperatures of water, it's reasonable to let $C$ run through at least $0$ to $100\text{.}$ Similarly it's reasonable to let $F$ run through at least $32$ to $212\text{.}$

2. Plot the $y$-intercept, which is at $(0,32)\text{.}$

3. Starting at the $y$-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 4.5.12 indicates, such slope triangles are too tiny. Since $\frac{9}{5}=\frac{90}{50}\text{,}$ we can try a “run” of $50$ and a rise of $90\text{.}$

###### Example4.5.13

Plot $y=-\frac{2}{3}x+10$ and $y=3x+5\text{.}$ These plots follow the approach from the previous example, but there is no context to the equation.

### Subsection4.5.3Writing 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 it's $y$-intercept.

### Subsection4.5.4Writing 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.

###### Example4.5.19

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

Solution

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 (4.4.3) from Section 4.4. 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}\\ \amp=-\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 $b=5\text{.}$ (We're only able to make this conclusion because this point has $0$ for its $x$-coordinate.) So, our equation is

\begin{equation*} y=-\frac{5}{4}x+5\text{.} \end{equation*}
###### Example4.5.20

Find the slope-intercept form of the equation of the line that passes through the points $(-8,15)$ and $(4,6)\text{.}$

Solution

The first step is always to find the slope between our two points: $(\overset{x_1}{-8},\overset{y_1}{15})$ and $(\overset{x_2}{4},\overset{y_2}{6})\text{.}$ Using the slope formula (4.4.3) again, we have:

\begin{align*} m\amp=\frac{y_2-y_1}{x_2-x_1}\\ \amp=\frac{\substitute{6}-\substitute{15}}{\substitute{4}-\substitute{(-8)}}\\ \amp=\frac{-9}{12}\\ \amp=-\frac{3}{4} \end{align*}

Now that we have the slope, we can write $y=-\frac{3}{4}x+b\text{.}$ Unlike in Example 4.5.19, 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 $(4,6)$ to plug in.

In conclusion, the equation for which we were searching is $y=-\frac{3}{4}x+9\text{.}$ 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.

###### Example4.5.21

Find the slope-intercept form of the equation of the line that passes through the points $(-3,\frac{9}{2})$ and $(4,-\frac{4}{3})\text{.}$

Solution

This example has fractions, but the process is the same: fractions are just numbers after all. First find the slope through our points: $\left(-3,\frac{9}{2}\right)$ and $\left(4,-\frac{4}{3}\right)\text{.}$ For this problem, we choose to do all of our algebra with improper fractions as it often simplifies the process.

\begin{align*} m\amp=\frac{y_2-y_1}{x_2-x_1}\\ \amp=\frac{\substitute{-\frac{4}{3}}-\substitute{\frac{9}{2}}}{\substitute{4}-\substitute{(-3)}}\\ \amp=\frac{-\frac{4}{3}\multiplyright{\frac{2}{2}}-\frac{9}{2}\multiplyright{\frac{3}{3}}}{7}\\ \amp=\frac{-\frac{8}{6}-\frac{27}{6}}{7}\\ \amp=\frac{-\frac{35}{6}}{\frac{7}{1}}\\ \amp=-\frac{35}{6}\cdot\frac{1}{7}\\ \amp=-\frac{5}{6} \end{align*}

So far we have $y=-\frac{5}{6}x+b\text{.}$ Now we need to solve for $b$ since neither of the points given were the vertical intercept. To do this, we will choose one of the two points and plug it into our equation. We choose $\left(-3,\frac{9}{2}\right)\text{.}$

Lastly, we write our equation.

\begin{equation*} y=-\frac{5}{6}x+7 \end{equation*}

### Subsection4.5.5Modeling 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.

###### Example4.5.25

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.

1. How much is the fare if a trip is $5.3$ miles long?

2. With $\100$ available to you, how long a trip can you afford?

Solution

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 $5$ miles long, we substitute $x=5$ into the equation and we have:

\begin{align*} y\amp=3.85x+7.35\\ \amp=3.85(\substitute{5})+7.35\\ \amp=19.25+7.35\\ \amp=26.60 \end{align*}

And the $5$-mile ride will cost you about $\26.60\text{.}$ (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.

### SubsectionExercises

Exercises on Identifying Slope and $y$-Intercept

###### 1

Find the line’s slope and $y$-intercept.

A line has equation $y={6}x+2\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 2

Find the line’s slope and $y$-intercept.

A line has equation $y={7}x+8\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 3

Find the line’s slope and $y$-intercept.

A line has equation $y={-4}x - 6\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 4

Find the line’s slope and $y$-intercept.

A line has equation $y={-3}x - 10\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 5

Find the line’s slope and $y$-intercept.

A line has equation $y=x+8\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 6

Find the line’s slope and $y$-intercept.

A line has equation $y=x - 10\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 7

Find the line’s slope and $y$-intercept.

A line has equation $y=-x - 8\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 8

Find the line’s slope and $y$-intercept.

A line has equation $y=-x - 6\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 9

Find the line’s slope and $y$-intercept.

A line has equation $\displaystyle{ y= -\frac{4x}{5} +4 }\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 10

Find the line’s slope and $y$-intercept.

A line has equation $\displaystyle{ y= -\frac{4x}{9} - 9 }\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 11

Find the line’s slope and $y$-intercept.

A line has equation $\displaystyle{ y= \frac{x}{6} +3 }\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 12

Find the line’s slope and $y$-intercept.

A line has equation $\displaystyle{ y= \frac{x}{8} - 4 }\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 13

Find the line’s slope and $y$-intercept.

A line has equation $\displaystyle{ y= 9 +{9}x }\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 14

Find the line’s slope and $y$-intercept.

A line has equation $\displaystyle{ y= 2 +{10}x }\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 15

Find the line’s slope and $y$-intercept.

A line has equation $\displaystyle{ y= 1 -x }\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 16

Find the line’s slope and $y$-intercept.

A line has equation $\displaystyle{ y= 2 -x }\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

Exercises on Graphing Lines in Slope-Intercept Form

###### 17

Graph the equation $y=4x\text{.}$

###### 18

Graph the equation $y=5x\text{.}$

###### 19

Graph the equation $y=-3x\text{.}$

###### 20

Graph the equation $y=-2x\text{.}$

###### 21

Graph the equation $y=\frac{5}{2}x\text{.}$

###### 22

Graph the equation $y=\frac{1}{4}x\text{.}$

###### 23

Graph the equation $y=-\frac{1}{3}x\text{.}$

###### 24

Graph the equation $y=-\frac{5}{4}x\text{.}$

###### 25

Graph the equation $y=5x+2\text{.}$

###### 26

Graph the equation $y=3x+6\text{.}$

###### 27

Graph the equation $y=-4x+3\text{.}$

###### 28

Graph the equation $y=-2x+5\text{.}$

###### 29

Graph the equation $y=x-4\text{.}$

###### 30

Graph the equation $y=x+2\text{.}$

###### 31

Graph the equation $y=-x+3\text{.}$

###### 32

Graph the equation $y=-x-5\text{.}$

###### 33

Graph the equation $y=\frac{2}{3}x+4\text{.}$

###### 34

Graph the equation $y=\frac{3}{2}x-5\text{.}$

###### 35

Graph the equation $y=-\frac{3}{5}x-1\text{.}$

###### 36

Graph the equation $y=-\frac{1}{5}x+1\text{.}$

Exercises on Writing Slope-Intercept Form Equation from the Graph

###### 37

A line’s graph is given.

This line’s slope-intercept equation is

###### 38

A line’s graph is given.

This line’s slope-intercept equation is

###### 39

A line’s graph is given.

This line’s slope-intercept equation is

###### 40

A line’s graph is given.

This line’s slope-intercept equation is

###### 41

A line’s graph is given.

This line’s slope-intercept equation is

###### 42

A line’s graph is given.

This line’s slope-intercept equation is

###### 43

A line’s graph is given.

This line’s slope-intercept equation is

###### 44

A line’s graph is given.

This line’s slope-intercept equation is

Exercises on Writing Slope-Intercept Form Given Two Points

###### 45

A line passes through the points $(2,12)$ and $(5,18)\text{.}$ Find this line’s equation in slope-intercept form.

This line’s slope-intercept equation is .

###### 46

A line passes through the points $(4,13)$ and $(2,9)\text{.}$ Find this line’s equation in slope-intercept form.

This line’s slope-intercept equation is .

###### 47

A line passes through the points $(-4,7)$ and $(-1,-5)\text{.}$ Find this line’s equation in slope-intercept form.

This line’s slope-intercept equation is .

###### 48

A line passes through the points $(1,1)$ and $(-4,21)\text{.}$ Find this line’s equation in slope-intercept form.

This line’s slope-intercept equation is .

###### 49

A line passes through the points $(-1,2)$ and $(-5,6)\text{.}$ Find this line’s equation in slope-intercept form.

This line’s slope-intercept equation is .

###### 50

A line passes through the points $(-5,9)$ and $(-1,5)\text{.}$ Find this line’s equation in slope-intercept form.

This line’s slope-intercept equation is .

###### 51

A line passes through the points $(-7,{1})$ and $(-21,{-15})\text{.}$ Find this line’s equation in slope-intercept form.

This line’s slope-intercept equation is .

###### 52

A line passes through the points $(-12,{-24})$ and $(-4,{-6})\text{.}$ Find this line’s equation in slope-intercept form.

This line’s slope-intercept equation is .

###### 53

A line passes through the points $(18,{5})$ and $(-9,{8})\text{.}$ Find this line’s equation in slope-intercept form.

This line’s slope-intercept equation is .

###### 54

A line passes through the points $(0,{1})$ and $(-14,{5})\text{.}$ Find this line’s equation in slope-intercept form.

This line’s slope-intercept equation is .

Applications

###### 55

A gym charges members ${\25}$ for a registration fee, and then ${\26}$ per month. You became a member some time ago, and now you have paid a total of ${\389}$ to the gym. How many months have passed since you joined the gym?

months have passed since you joined the gym.

###### 56

Your cell phone company charges a ${\17}$ monthly fee, plus ${\0.19}$ per minute of talk time. One month your cell phone bill was ${\110.10}\text{.}$ How many minutes did you spend talking on the phone that month?

You spent talking on the phone that month.

###### 57

A school purchased a batch of T-shirts from a company. The company charged ${\6}$ per T-shirt, and gave the school a ${\80}$ rebate. If the school had a net expense of ${\2{,}140}$ from the purchase, how many T-shirts did the school buy?

The school purchased T-shirts.

###### 58

Dave hired a face-painter for a birthday party. The painter charged a flat fee of ${\80}\text{,}$ and then charged ${\3.50}$ per person. In the end, Dave paid a total of ${\167.50}\text{.}$ How many people used the face-painter’s service?

people used the face-painter’s service.

###### 59

A certain country has $570.24$ million acres of forest. Every year, the country loses $7.92$ million acres of forest mainly due to deforestation for farming purposes. If this situation continues at this pace, how many years later will the country have only $300.96$ million acres of forest left? (Use an equation to solve this problem.)

After years, this country would have $300.96$ million acres of forest left.

###### 60

Ross has ${\86}$ in his piggy bank. He plans to purchase some Pokemon cards, which costs ${\2.55}$ each. He plans to save ${\40.10}$ to purchase another toy. At most how many Pokemon cards can he purchase?

Write an equation to solve this problem.

Ross can purchase at most Pokemon cards.

###### 61

By your cell phone contract, you pay a monthly fee plus some money for each minute you use the phone during the month. In one month, you spent $210$ minutes on the phone, and paid ${\25.65}\text{.}$ In another month, you spent $360$ minutes on the phone, and paid ${\35.40}\text{.}$

Let $x$ be the number of minutes you talk over the phone in a month, and let $y$ be your cell phone bill, in dollars, for that month. Use a linear equation to model your monthly bill based on the number of minutes you talk over the phone.

1. This linear model’s slope-intercept equation is .

2. If you spent $120$ minutes over the phone in a month, you would pay .

3. If in a month, you paid ${\40.60}$ of cell phone bill, you must have spent minutes on the phone in that month.

###### 62

A company set aside a certain amount of money in the year 2000. The company spent exactly the same amount from that fund each year on perks for its employees. In $2003\text{,}$ there was still ${\888{,}000}$ left in the fund. In $2005\text{,}$ there was ${\848{,}000}$ left.

Let $x$ be the number of years since 2000, and let $y$ be the amount of money, in dollars, left in the fund that year. Use a linear equation to model the amount of money left in the fund after so many years.

1. The linear model’s slope-intercept equation is .

2. In the year $2009\text{,}$ there was left in the fund.

3. In the year , the fund will be empty.

###### 63

A biologist has been observing a tree’s height. $15$ months into the observation, the tree was $17$ feet tall. $20$ months into the observation, the tree was $17.6$ feet tall.

Let $x$ be the number of months passed since the observations started, and let $y$ be the tree’s height at that time, in feet. Use a linear equation to model the tree’s height as the number of months pass.

1. This line’s slope-intercept equation is .

2. $26$ months after the observations started, the tree would be feet in height.

3. months after the observation started, the tree would be $21.56$ feet tall.

###### 64

Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way.

Six minutes since the experiment started, the gas had a mass of $93.6$ grams.

Seventeen minutes since the experiment started, the gas had a mass of $65$ grams.

Let $x$ be the number of minutes that have passed since the experiment started, and let $y$ be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time.

1. This line’s slope-intercept equation is .

2. $31$ minutes after the experiment started, there would be grams of gas left.

3. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone.

###### 65

By your cell phone contract, you pay a monthly fee plus ${\0.04}$ for each minute you spend on the phone. In one month, you spent $270$ minutes over the phone, and had a bill totaling ${\28.80}\text{.}$

Let $x$ be the number of minutes you spend on the phone in a month, and let $y$ be your total cell phone bill for that month, in dollars. Use a linear equation to model your monthly bill based on the number of minutes you spend on the phone.

1. This line’s slope-intercept equation is .

2. If you spend $120$ minutes on the phone in a month, you would be billed .

3. If your bill was ${\35.60}$ one month, you must have spent minutes on the phone in that month.

###### 66

A company set aside a certain amount of money in the year 2000. The company spent exactly ${\33{,}000}$ from that fund each year on perks for its employees. In $2002\text{,}$ there was still ${\726{,}000}$ left in the fund.

Let $x$ be the number of years since 2000, and let $y$ be the amount of money, in dollars, left in the fund that year. Use a linear equation to model the amount of money left in the fund after so many years.

1. The linear model’s slope-intercept equation is .

2. In the year $2009\text{,}$ there was left in the fund.

3. In the year , the fund will be empty.

###### 67

A biologist has been observing a tree’s height. This type of tree typically grows by $0.21$ feet each month. Thirteen months into the observation, the tree was $14.73$ feet tall.

Let $x$ be the number of months passed since the observations started, and let $y$ be the tree’s height at that time, in feet. Use a linear equation to model the tree’s height as the number of months pass.

1. This line’s slope-intercept equation is .

2. $25$ months after the observations started, the tree would be feet in height.

3. months after the observation started, the tree would be $23.13$ feet tall.

###### 68

Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Its mass is leaking by $8$ grams each minute. Ten minutes since the experiment started, the remaining gas had a mass of $296$ grams.

Let $x$ be the number of minutes that have passed since the experiment started, and let $y$ be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time.

1. This line’s slope-intercept equation is .

2. $31$ minutes after the experiment started, there would be grams of gas left.

3. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone.