## Algebra 1: Common Core (15th Edition)

The x-intercept is $10$. The y-intercept is $-\frac{10}{3}$.
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: $m = \frac{-2 - (-5)}{4 - (-5)}$ Simplify by adding or subtracting in the numerator and denominator: $m = \frac{3}{9}$ Simplify by dividing both the numerator and denominator by their greatest common factor, $3$: $m = \frac{1}{3}$ 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: $y - (-2) = \frac{1}{3}(x - 4)$ Simplify the right side of the equation: $y + 2 = \frac{1}{3}(x - 4)$ This is the point-slope formula of the equation. To find the x-intercept, we set $y$ equal to $0$: $0 + 2 = \frac{1}{3}(x - 4)$ Use the distributive property on the right side of the equation: $2 = \frac{1}{3}x - \frac{4}{3}$ Add $\frac{4}{3}$ to both sides of the equation to move constants to one side of the equation: $\frac{1}{3}x = 2 + \frac{4}{3}$ Add the fractions on the right side of the equation: $\frac{1}{3}x = \frac{10}{3}$ Divide each side by $\frac{1}{3}$ to solve for $x$: $x = \frac{30}{3}$ Divide both the numerator and denominator by their greatest common factor, $3$: $x = 10$ To find the y-intercept, we set $x$ equal to $0$: $y + 2 = \frac{1}{3}(0 - 4)$ Evaluate parentheses first: $y + 2 = \frac{1}{3}(-4)$ Multiply on the right side of the equation: $y + 2 = -\frac{4}{3}$ Subtract $2$ from each side of the equation: $y = -\frac{4}{3} - 2$ Subtract to solve for $y$: $y = -\frac{10}{3}$ The x-intercept is $10$. The y-intercept is $-\frac{10}{3}$.