Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.2 Representing Functions - 1.2 Exercises - Page 21: 14


The function is $$f(x)=x-5.$$ The graph is in the figure below

Work Step by Step

The equation of the linear function passing through points $P(x_1,y_1)$ and $Q(x_2,y_2)$ is given by $$y=\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1.$$ Using the data given in the problem $$x_1=2\quad y_1=-3,\quad x_2=5,\quad y_2=0$$ we get $$y=\frac{0-(-3)}{5-2}(x-2)+(-3)=x-2-3=x-5$$ so our linear function is $$f(x)=x-5.$$ Drawing the line through points $(2,-3)$ and $(5,0)$ we get the required graph
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