1. **Find the equation of the line through (13, 5) with slope $m = -2$.** The slope-intercept form is given by: $$y = mx + b$$ where $m$ is the slope and $b$ is the y-intercept. 2. Substitute the point $(13, 5)$ and slope $m = -2$ into the equation to find $b$: $$5 = -2(13) + b$$ $$5 = -26 + b$$ Add 26 to both sides: $$5 + 26 = \cancel{-26} + b + 26$$ $$31 = b$$ 3. So the equation is: $$y = -2x + 31$$ **Answer for first problem: Option B.** --- 4. **Find the horizontal asymptote of** $$f(x) = \frac{6x^2 + 6}{6x^2 - 6}$$ For rational functions, horizontal asymptotes depend on the degrees of numerator and denominator. - Both numerator and denominator have degree 2. - The horizontal asymptote is the ratio of leading coefficients: $$y = \frac{6}{6} = 1$$ **Answer for second problem: Option A.** --- 5. **Find the equation of the line through (0, -4) with slope $m = \frac{7}{3}$.** Using slope-intercept form: $$y = mx + b$$ Since the point is $(0, -4)$, the y-intercept $b = -4$. So the equation is: $$y = \frac{7}{3}x - 4$$ **Answer for third problem: Option B.**