What if the linear function \begin{align*}W(g)\end{align*} represented a family's monthly water bill, with \begin{align*}g\end{align*} as the number of gallons of water used. If you knew what the function was, could you find \begin{align*}W(25)\end{align*}? How about if you knew the slope of the function and the value of \begin{align*}W(25)\end{align*}? Could you determine what the function was? Suppose you knew the values of \begin{align*}W(25)\end{align*} and \begin{align*}W(50)\end{align*}. Could you determine the function in this case? After completing this Concept, you'll be able to perform tasks like these.

### Guidance

Remember that a linear function has the form \begin{align*}f(x)=mx+b\end{align*}. Here \begin{align*}f(x)\end{align*} represents the \begin{align*}y\end{align*} values of the equation or the graph. So \begin{align*}y=f(x)\end{align*} and they are often used interchangeably. Using the functional notation in an equation often provides you with more information.

For instance, the expression \begin{align*}f(x)=mx+b\end{align*} shows clearly that \begin{align*}x\end{align*} is the independent variable because you **substitute** values of \begin{align*}x\end{align*} into the function and perform a series of operations on the value of \begin{align*}x\end{align*} in order to calculate the values of the dependent variable, \begin{align*}y\end{align*}.

In this case when you substitute \begin{align*}x\end{align*} into the function, the function tells you to multiply it by \begin{align*}m\end{align*} and then add \begin{align*}b\end{align*} to the result. This process generates all the values of \begin{align*}y\end{align*} you need.

#### Example A

Consider the function \begin{align*}f(x)=3x-4.\end{align*} *Find* \begin{align*}f(2), f(0),\end{align*} *and* \begin{align*}f(-1)\end{align*}.

**Solution:**

Each number in parentheses is a value of \begin{align*}x\end{align*} that you need to substitute into the equation of the function.

\begin{align*}f(2)=2; f(0)=-4; \ and \ f(-1)=-7\end{align*}

Function notation tells you much more than the value of the independent variable. It also indicates a point on the graph. For example, in the above example, \begin{align*}f(-1)=-7\end{align*}. This means the ordered pair (–1, –7) is a solution to \begin{align*}f(x)=3x-4\end{align*} and appears on the graphed line. You can use this information to write an equation for a function.

#### Example B

*Write an equation for a line with \begin{align*}m=3.5\end{align*} and \begin{align*}f(-2)=1\end{align*}*.

**Solution:**

You know the slope, and you know a point on the graph, (–2, 1). Using the methods presented in this Concept, write the equation for the line.

Begin with slope-intercept form.

\begin{align*}&& y& =mx+b\\ \text{Substitute the value for the slope.} && y& =3.5x+b\\ \text{Use the ordered pair to solve for} \ b. && 1& =3.5(-2)+b\\ && b& =8\\ \text{Rewrite the equation.} && y& =3.5x+8 \\ \text{or} && f(x)& =3.5x+8\end{align*}

#### Example C

*Write an equation for a line with \begin{align*}f(-1)=2\end{align*} and \begin{align*}f(5)=20\end{align*}*.

**Solution:**

You know two points on the graph. Using the methods presented in the previous Concept, write the equation for the line. First, you must find the slope:

\begin{align*}m=\frac{y_2-y_1}{x_2-x_1}=\frac{20-2}{5-(-1)}=\frac{18}{6}=3\end{align*}.

Now use the slope-intercept form:

\begin{align*}&& y& =mx+b\\ \text{Substitute the value for the slope.} && y& =3x+b\\ \text{Use the ordered pair to solve for} \ b. && 2& =3(-1)+b\\ && b& =5\\ \text{Rewrite the equation.} && y& =3x+5\\ \text{or} && f(x)& =3x+5\end{align*}

### Video Review

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### Guided Practice

*Write an equation for a line with \begin{align*}f(0)=2\end{align*} and \begin{align*}f(3)=-4\end{align*}* and use it to find \begin{align*}f(-5), f(2), f(0), \text{ and } f(z)\end{align*}.

**Solution:**

Notice that the first point given as an input value is 0, and the output is 2, which means the point is (0,2). This is the \begin{align*}y\end{align*}-intercept. So, all we have to do is find the slope and then plug both values into the slope-intercept form:

\begin{align*}m=\frac{y_2-y_1}{x_2-x_1}=\frac{-4-2}{3-0}=\frac{-6}{3}=-2\end{align*}.

Now use the slope-intercept form.

\begin{align*}&& y& =mx+b\\ \text{Substitute the value for the slope.} && y& =-2x+b\\ \text{Substitute the value for the y-intercept} && y& =-2x+2\\ \text{or} && f(x)& =-2x+2\end{align*}

Now we find the values of \begin{align*}f(-5), f(2), f(0), \text{ and } f(z)\end{align*} for \begin{align*} f(x) =-2x+2 \end{align*}.

\begin{align*} f(-5) =-2(-5)+2=10+2=12 \end{align*}

\begin{align*} f(2) =-2(2)+2=-2 \end{align*}

\begin{align*} f(0) =-2(0)+2=0 \end{align*}

\begin{align*} f(z) =-2z+2 \end{align*}

### Explore More

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Linear Equations in Slope-Intercept Form (14:58)

- Consider the function \begin{align*}f(x)=-2x-3.\end{align*}
*Find*\begin{align*}f(-3), f(0),\end{align*}*and*\begin{align*}f(5)\end{align*}. - Consider the function \begin{align*}f(x)=\frac{2}{3}x+10.\end{align*}
*Find*\begin{align*}f(-9), f(0),\end{align*}*and*\begin{align*}f(9)\end{align*}.

In 3–10, find the equation of the linear function in slope–intercept form.

- \begin{align*}m=5, f(0)=-3\end{align*}
- \begin{align*}m=-2\end{align*}, \begin{align*}f(0)=5\end{align*}
- \begin{align*}m=-7, f(2)=-1\end{align*}
- \begin{align*}m=\frac{1}{3}, f(-1)=\frac{2}{3}\end{align*}
- \begin{align*}m=4.2, f(-3)=7.1\end{align*}
- \begin{align*}f\left (\frac{1}{4}\right )=\frac{3}{4}, f(0)=\frac{5}{4}\end{align*}
- \begin{align*}f(1.5)=-3, f(-1)=2\end{align*}
- \begin{align*}f(-1)=1\end{align*}, \begin{align*}f(1)=-1\end{align*}

**Mixed Review**

- Translate into a sentence: \begin{align*}4(j+2)=400\end{align*}.
- Evaluate \begin{align*}0.45 \cdot 0.25-24 \div \frac{1}{4}\end{align*}.
- The formula to convert Fahrenheit to Celsius is \begin{align*}C(F)=\frac{F-32}{1.8}\end{align*}. What is the Celsius equivalent to \begin{align*}35^\circ F\end{align*}?
- Find the rate of change: The diver dove 120 meters in 3 minutes.
- What percent of 87.4 is 106?
- Find the percent of change: The original price was $25.00. The new price is $40.63.
- Solve for \begin{align*}w: \ 606=0.045(w-4000)+0.07w\end{align*}.

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 5.3.