Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 5 - Linear Functions - 5-5 Standard Form - Practice and Problem-Solving Exercises - Page 327: 54


The x-intercept is $\frac{-53}{5}$ The y-intercept is $\frac{53}{3}$

Work Step by Step

To find the x-intercept and the y-intercept of the line, we first need to find the equation of the line. We can use the two points given to formulate the point-slope form. Let's first find the slope: $m=\frac{y_2-y_1}{x_2-x_1}$, where $m$ is the slope and $(x_1,y_1)$ and $(x_2,y_2)$ are points on the line. Let's plug in our two points into this formula: $m=\frac{11-6}{-4-(-7)}=\frac{5}{3}$ Since we have the slope and two points, we can use the point-slope form, which is given by the following formula: $y-y_1=m(x-x_1)$ Let's plug in the slope and a point into this formula: $y-11=\frac{5}{3}(x-(-4))$ $y-11=\frac{5}{3}(x+4)$ This is the point-slope formula of the equation. To find the x-intercept, we set y equal to 0: $0-11=\frac{5}{3}(x+4)$ Use the distributive property on the right side of the equation: $\frac{5}{3}x+\frac{20}{3}=-11$ $x=\frac{-53}{5}$ To find the y-intercept, we set x equal to 0: $y-11=\frac{5}{3}(0+4)$ $y-11=\frac{20}{3}$ $y=\frac{53}{3}$ The x-intercept is $\frac{-53}{5}$. The y-intercept is $\frac{53}{3}$.
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