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: 52


x-intercept = $-2$ y-intercept = $-2$

Work Step by Step

To find the x-intercept and 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: $m = \frac{-5 - 4}{3 - (-6)}$ Simplify by adding or subtracting in the numerator and denominator: $m = \frac{-9}{9}$ Simplify by dividing both the numerator and denominator by their greatest common factor, $9$: $m = -1$ 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: $y - 4 = -1(x - (-6))$ Simplify the right side of the equation: $y - 4 = -1(x + 6)$ Use the distributive property on the right side of the equation: $y - 4 = -x - 6$ This would be the point-slope form of the equation, but we want the slope-intercept form, so we need to isolate $y$ on the left side of the equation by adding $4$ to both sides of the equation: $y = -x - 2$ To find the x-intercept, we set $y$ equal to $0$: $0 = -x - 2$ Add $x$ to each side of the equation to move the variable to the left side of the equation: $x = -2$ To find the y-intercept, we set $x$ equal to $0$: $y = -(0) - 2$ Subtract to solve: $y = -2$ The x-intercept is $-2$. The y-intercept is $-2$.
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