## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 1 - Precalculus Review - 1.2 Linear and Quadratic Functions - Exercises - Page 18: 3

#### Answer

Slope: $m=-\dfrac{4}{9}$ $x$-intercept: $\Big(\dfrac{3}{4},0\Big)$ $y$-intercept: $\Big(0,\dfrac{1}{3}\Big)$

#### Work Step by Step

$4x+9y=3$ Express this equation inslope-intercept form by solving it for $y$: $9y=-4x+3$ $y=-\dfrac{4}{9}x+\dfrac{3}{9}$ $y=-\dfrac{4}{9}x+\dfrac{1}{3}$ This line is now in slope-intercept from, which is $y=mx+b$, where $m$ is the slope of the line and $b$ is its $y$-intercept. Comparing the given line to the slope-intercept form, it can be seen that $m=-\dfrac{4}{9}$ and $b=\dfrac{1}{3}$ The slope of the given line is $m=-\dfrac{4}{9}$ and its $y$-intercept is the point $\Big(0,\dfrac{1}{3}\Big)$ To find the $x$-intercept, set $y$ equal to $0$ in the original equation and solve for $x$: $4x+9(0)=3$ $4x=3$ $x=\dfrac{3}{4}$ The $x$-intercept of the given line is the point $\Big(\dfrac{3}{4},0\Big)$

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