Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter F - Foundations: A Prelude to Functions - Section F.3 Lines - F.3 Assess Your Understanding - Page 31: 90

Answer

The $y$-intercept is $0$. The slope is $-\dfrac{3}{2}$. See the graph below.

Work Step by Step

First, we have to get the equation's slope-intercept form, that is $y=ax+b$. Isolate $y$ to obtain: \begin{align*} 3x+2y-3x&=0-3x\\ 2y&=-3x\\ \frac{2y}{2}&=\frac{-3x}{2}\\ y&=-\frac{3}{2}x \end{align*} This form of the equation is the slope-intercept form. In this form, the slope of the line equals to the coefficient of $x$ (which is $m$) and the $y$-intercept equals to the constant $b$. Therefore in the equation $=-\dfrac{3}{2}x$: The $y$-intercept is $0$. The slope is $-\dfrac{3}{2}$. In order to graph the line, we have to sketch the $y$-intercept, that is $(0, 0)$. As the slope is $-\frac{3}{2}$, we can find another point that we can also sketch. The slope is the change in $y$ for every $1$ unit change of $x$. Thus, a slope of $-\frac{3}{2}$ means a $1$-unit increase in $x$ will result to a $-\frac{3}{2}$-unit increase (or $\frac{3}{2}$-unit decrease) in $y$. This is equivalent to a $3$ unit decrease in $y$ for a $2$-unit increase in $x$. Using $(0, 0)$ as the starting point and a slope of $-\frac{3}{2}$, the coordinates of another point on the line would be: $(0+2, 0-3)=(2, -3)$ Plot the two points then connect them using a straigiht line. Refer to the graph above,
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