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 \(90\text{.}\)

  4. Connect your points, use arrowheads, and label the equation.

Figure4.5.12Graphing \(F=\frac{9}{5}C+32\)
in-context