1. **State the problem:** Find the x- and y-intercepts of the line given by the equation $$y = \frac{9}{5}x + 32$$. 2. **Recall intercept definitions:** - The **y-intercept** is where the line crosses the y-axis, so $x=0$. - The **x-intercept** is where the line crosses the x-axis, so $y=0$. 3. **Find the y-intercept:** Substitute $x=0$ into the equation: $$y = \frac{9}{5} \times 0 + 32 = 32$$ So the y-intercept is at the point $(0, 32)$. 4. **Find the x-intercept:** Set $y=0$ and solve for $x$: $$0 = \frac{9}{5}x + 32$$ $$\frac{9}{5}x = -32$$ Multiply both sides by $\cancel{\frac{5}{9}}$ to isolate $x$: $$x = -32 \times \cancel{\frac{5}{9}}$$ $$x = -\frac{160}{9}$$ So the x-intercept is at the point $\left(-\frac{160}{9}, 0\right)$. 5. **Answer:** - x-intercept: $\left(-\frac{160}{9}, 0\right)$ - y-intercept: $(0, 32)$ 6. **Match with options:** This corresponds to option D. **Final answer:** x-intercept = $\left(-\frac{160}{9}, 0\right)$; y-intercept = $(0, 32)$.