
## SectionFinding Linear Functions

### SubsectionFinding a Linear Equation from a Graph

We can also use the slope-intercept form to find the equation of a line from its graph. First, we note the value of the $y$-intercept from the graph, and then we calculate the slope using two convenient points.

###### Example90

Find an equation for the line shown in Figure91.

Solution

The line crosses the $y$-axis at the point $(0, 3200)\text{,}$ so $b=3200\text{.}$ To calculate the slope of the line, we locate another point, say $(20, 6000)\text{,}$ and compute:

\begin{align*}m \amp = \frac{\Delta y}{\Delta x} = \frac{6000 - 3200}{20 - 0}\\ \amp =\frac{2800}{20}= 140 \end{align*} The slope-intercept form of the equation, with $m = 140$ and $b = 3200\text{,}$ is

\begin{equation*} y = 3200 + 140x\text{.} \end{equation*}
###### Example93

Find an equation for the line shown in Figure94.

Solution

The line crosses the $y$-axis at the point $(0, 80)\text{,}$ so $b=80\text{.}$ To calculate the slope of the line, we locate another point, say $(20,30)\text{,}$ and compute:

\begin{equation*} \begin{aligned}m \amp = \frac{\Delta y}{\Delta x} = \frac{30 - 80}{20 - 0}\\ \amp =\frac{-50}{20}= -\frac{5}{2} \end{aligned} \end{equation*}

The slope-intercept form of the equation is

\begin{equation*} y = 80 - \frac{5}{2}x\text{.} \end{equation*}

### SubsectionPoint-Slope Form

We can find an equation for a line if we know its slope and $y$-intercept. What if we do not know the $y$-intercept, but instead know some other point on the line? There is only one line that passes through a given point and has a given slope, so we should be able to find an equation for this line.

For example, we can graph the line of slope $\dfrac{-3}{4}$ that passes through the point $(1, -4)\text{.}$ We first plot the given point, $(1, -4)\text{,}$ as shown in Figure95.

Then we use the slope to find another point on the line. The slope is

\begin{equation*} m =\frac{-3}{4}= \frac{\Delta y}{\Delta x} \end{equation*}

so we move down $3$ units and then $4$ units to the right, starting from $(1, -4)\text{.}$ This brings us to the point $(5, -7)\text{.}$ We can then draw the line through these two points.

We can also find an equation for the line, as shown in Example96.

###### Example96

Find an equation for the line that passes through $(1, -4)$ and has slope $\dfrac{-3}{4}\text{.}$

Solution

We will use the formula for slope,

\begin{equation*} m = \frac{y_2 - y_1}{x_2 - x_1}. \end{equation*}

We substitute $\dfrac{-3}{4}$ for the slope, $m\text{,}$ and $(1, -4)$ for $(x_1, y_1)\text{.}$ For the second point, $(x_2, y_2)\text{,}$ we use the variable point $(x, y)\text{.}$ Substituting these values into the slope formula gives us

\begin{equation*} \frac{-3}{4}= \frac{y - (-4)}{x - 1}=\frac{y + 4}{x - 1}. \end{equation*}

To solve for $y$ we first multiply both sides by $x - 1\text{.}$ \begin{align*} \alert{(x-1)}\frac{-3}{4}\amp =\frac{y +4}{x - 1}\alert{(x-1)} \amp\amp\\ \frac{-3}{4}(x-1)\amp=y+4 \amp\amp\text{Apply the distributive law.}\\ \frac{-3}{4}x+\frac{3}{4}\amp=y+4 \amp\amp \text{Subtract 4 from both sides.}\\ \frac{-3}{4}x-\frac{13}{4}\amp=y \amp\amp \frac{3}{4}-4=\frac{3}{4}-\frac{16}{4}=\frac{-13}{4} \end{align*} An equation of the line is $y=\dfrac{-13}{4}-\dfrac{3}{4}x.$

When we use the slope formula in this way to find an equation of a line, we substitute a variable point $(x, y)$ for the second point. This version of the formula,

\begin{equation*} m = \frac{y - y_1}{x - x_1} \end{equation*}

is called the point-slope form for a linear equation. It is sometimes stated in another form obtained by clearing the fraction to get \begin{align*} \alert{(x-x_1)}m\amp =\frac{y -y_1}{x - x_1}\alert{(x-x_1)} \amp\amp \text{Multiply both sides by }(x - x_1)\\ (x - x_1) m\amp=y-y_1 \amp\amp\text{Clear fractions and solve for }y\text{.}\\ y\amp=y_1+m(x-x_1). \end{align*}

###### Point-Slope Form

An equation of the line that passes through the point $(x_1, y_1)$ and has slope $m$ is

\begin{equation*} y= y_1 + m(x- x_1). \end{equation*}
###### Example97

Use the point-slope form to find an equation of the line that passes through the point $(-3, 5)$ and has slope $-1.4\text{.}$

Solution

Subsituting $m=-1.4$ and $(-3,5)$ for $(x_1,y_1)$ into the point-slope form of a line, we have

\begin{equation*} \begin{aligned} y\amp=5-1.4(x-(-3)) \\ y\amp=0.8-1.4x \end{aligned} \end{equation*}

The point-slope form is useful for modeling linear functions when we do not know the initial value but do know some other point on the line.

###### Example98

Under a proposed graduated income tax system, single taxpayers would owe $1500 plus 20% of the amount of their income over$13,000. (For example, if your income is $18,000, you would pay$1500 plus 20% of $5000.) 1. Complete the table of values for the tax, $T\text{,}$ on various incomes, $I\text{.}$  $I$ $15,000$ $20,000$ $22,000$ $T$ 2. Write a linear equation in point-slope form for the tax, $T\text{,}$ on an income $I\text{.}$ 3. Write the equation in slope-intercept form. Solution 1. On an income of$15,000, the amount of income over $13,000 is$15,000 - $13,000 =$2000, so you would pay $1500 plus 20% of$2000, or

\begin{equation*} T = 1500 + 0.20(2000) = 1900. \end{equation*}

You can compute the other function values in the same way.

 $I$ $15,000$ $20,000$ $22,000$ $T$ $1900$ $2900$ $3300$
2. On an income of $I\text{,}$ the amount of income over $13,000 is $I - \13,000\text{,}$ so you would pay$1500 plus 20% of $I - 13,000\text{,}$ or
\begin{equation*} T = 1500 + 0.20 (I - 13,000). \end{equation*}
3. Simplify the right side of the equation to get \begin{align*} T \amp = 1500 + 0.20I - 2600\\ T \amp = -1100 + 0.20I. \end{align*}
###### Example99

A healthy weight for a young woman of average height, 64 inches, is 120 pounds. To calculate a healthy weight for a woman taller than 64 inches, add 5 pounds for each inch of height over 64.

1. Write a linear equation in point-slope form for the healthy weight, $W\text{,}$ for a woman of height, $H\text{,}$ in inches.

2. Write the equation in slope-intercept form.

Solution
1. $W = 120 + 5(H - 64)$

2. $W = -200 + 5H$

If you would like more information on linear equations, refer to the following two sections from Intermediate Algebra: