3.4 The Point-Slope Form of a Line
3.4 The Point-Slope Form of a Line
3.4 The Point-Slope Form of a Line
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302 Chapter 3 <strong>Line</strong>ar Functions<br />
Version: Fall 2007<br />
120<br />
C<br />
(32, 0)<br />
(212, 100)<br />
F<br />
250<br />
Figure 6. Plotting Celsius temperature versus Fahrenheit<br />
temperature.<br />
y − 0 = 5<br />
(x − 32). (20)<br />
9<br />
However, our dependent axis is labeled C, not y, and our independent axis is labeled<br />
F , not x. So, we must replace y and x in equation (20) with C and F , respectively,<br />
obtaining<br />
C = 5<br />
(F − 32). (21)<br />
9<br />
This result in equation (21) expresses the Celsius temperature as a function <strong>of</strong> the<br />
Fahrenheit temperature. Alternatively, we could also use function notation and write<br />
C(F ) = 5<br />
(F − 32).<br />
9<br />
Suppose that we know that the Fahrenheit temperature outside is 80 ◦ F and we<br />
wish to express this using the Celsius scale. To do so, we simply evaluate C(80), as in<br />
C(80) = 5<br />
(80 − 32) ≈ 26.6.<br />
9<br />
Hence, the Celsius temperature is approximately 26.6 ◦ C.<br />
On the other hand, suppose that we know the Celsius temperature on a metal ro<strong>of</strong><br />
is 80 ◦ C and we wish to find the Fahrenheit temperature. To do so, we need to solve<br />
for F , or equivalently,<br />
Multiply both sides by 9 to obtain<br />
C(F ) = 80<br />
5<br />
(F − 32) = 80.<br />
9<br />
5(F − 32) = 720,
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