Algebra 1: Common Core (15th Edition)

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

Chapter 5 - Linear Functions - Chapter Review - Page 355: 20


$y = \frac{2}{3}x + 1$

Work Step by Step

The first thing we want to do in this problem is figure out two points on this graph. Once we have these two points, we can find the slope. We can see that the graph passes through the point $(3, 3)$ and the point $(-3, -1)$. Let's use the formula to find the slope $m$ given two points: $m = \frac{y_2 - y_1}{x_2 - x_1}$ Let's plug in the values into this formula: $m = \frac{-1 - 3}{-3 - 3}$ Subtract the numerator and denominator to simplify: $m = \frac{-4}{-6}$ Divide the numerator and denominator by their greatest common factor, which is $-2$: $m = \frac{2}{3}$ Now that we have the slope, we can use one of the points and plug these values into the point-slope equation, which is given by the formula: $y - y_1 = m(x - x_1)$ Let's plug in the points and slope into the formula: $y - 3 = \frac{2}{3}(x - 3)$ To change this equation into point-intercept form, we need to isolate $y$. Use distribution to simplify: $y - 3 = \frac{2}{3}x + \frac{2}{3}(-3)$ Multiply to simplify: $y - 3 = \frac{2}{3}x + \frac{-6}{3}$ Simplify the fraction by dividing the numerator by the denominator: $y - 3 = \frac{2}{3}x - 2$ To isolate $y$, we add $3$ to each side of the equation: $y = \frac{2}{3}x - 2 + 3$ Add to simplify: $y = \frac{2}{3}x + 1$ Now, we have the equation of the line in slope-intercept form.
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