Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.2 Linear and Quadratic Functions - Exercises - Page 18: 25


(a): $c= \frac{-1}{4}$ (b): $c=-2$ (c): no value of $c$ (d): $c=0$

Work Step by Step

(a): $slope = \frac{-1}{c} = 4$, which implies that $c= \frac{-1}{4}$. (b): The equation $x+cy=1$ must satisfy that the line that passes through $(3,1)$; that is, $3+c=1 \to c=-2$. (c): In order for the line to be horizontal, we must have the slope $\frac{-1}{c} = 0$, but solving for $c$ would require division by zero, which is not allowed. Thus, there is no value of $c$ in this case. (d): For the line to be vertical means that the slope $\frac{-1}{c} \to \infty$ or $\frac{-1}{c} \to - \infty$, which means that the value of $c$ must be $0$ in this case.
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