Section4.5Slope-Intercept Form
Ā¶In this section, we will explore one of the āstandardā ways to write the equation of a line. It's known as slope-intercept form.
Subsection4.5.1Slope-Intercept Definition
Recall ExampleĀ 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 SectionĀ 4.4. We also saw in FigureĀ 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. FigureĀ 4.5.2 illustrates this in the abstract.
What else is there?
Can you think of a type of straight line that does not have a notion of slope? Or that does not cross the \(y\)-axis somewhere?
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 \(y\)-interecept \((0,b)\) where \(b\) represents the position of the \(y\)-intercept on the \(y\)-axis.
Exercise4.5.3
Use FigureĀ 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
\begin{equation} y=mx+b\tag{4.5.1} \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.
Exercise4.5.5
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 formĀ (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 SectionĀ 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 \(y=3x-4\text{.}\)
Exercise4.5.9
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 SectionĀ 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 \((0,32)\text{.}\)
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 \(212\text{.}\)
Plot the \(y\)-intercept, which is at \((0,32)\text{.}\)
Starting at the \(y\)-intercept, use slope triangles to reach the next point. Since our slope is \(\frac{9}{5}\text{,}\) that suggests a ārunā of \(5\) and a rise of \(9\) might work. But as FigureĀ 4.5.12 indicates, such slope triangles are too tiny. Since \(\frac{9}{5}=\frac{90}{50}\text{,}\) we can try a ārunā of \(50\) and a rise of \(90\text{.}\)
Connect your points, use arrowheads, and label the equation.
Example4.5.13
Plot \(y=-\frac{2}{3}x+10\) and \(y=3x+5\text{.}\) These plots follow the approach form the previous example, but there is no context to the equation.
Subsection4.5.3Writing a Slope-Intercept Equation
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.
Exercise4.5.17
Exercise4.5.18
Subsection4.5.4Modeling 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.
Example4.5.20
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 30āÆmph, 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 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 āabout,ā because this was all assuming you average 30āÆmph.)
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.
Exercise4.5.21
Subsection4.5.5Exercises
Exercises on Identifying Slope and \(y\)-Intercept
Exercises on Graphing Lines in Slope-Intercept Form
7
Graph the equation \(y=4x\text{.}\)
8
Graph the equation \(y=5x\text{.}\)
9
Graph the equation \(y=-3x\text{.}\)
10
Graph the equation \(y=-2x\text{.}\)
11
Graph the equation \(y=\frac{5}{2}x\text{.}\)
12
Graph the equation \(y=\frac{1}{4}x\text{.}\)
13
Graph the equation \(y=-\frac{1}{3}x\text{.}\)
14
Graph the equation \(y=-\frac{5}{4}x\text{.}\)
15
Graph the equation \(y=5x+2\text{.}\)
16
Graph the equation \(y=3x+6\text{.}\)
17
Graph the equation \(y=-4x+3\text{.}\)
18
Graph the equation \(y=-2x+5\text{.}\)
19
Graph the equation \(y=x-4\text{.}\)
20
Graph the equation \(y=x+2\text{.}\)
21
Graph the equation \(y=-x+3\text{.}\)
22
Graph the equation \(y=-x-5\text{.}\)
23
Graph the equation \(y=\frac{2}{3}x+4\text{.}\)
24
Graph the equation \(y=\frac{3}{2}x-5\text{.}\)
25
Graph the equation \(y=-\frac{3}{5}x-1\text{.}\)
26
Graph the equation \(y=-\frac{1}{5}x+1\text{.}\)
Exercises on Writing Slope-Intercept Form Equation from the Graph
Exercises on Slope-Intercept Form Equation Applications