1. **State the problem:** Find where the function $f(x) = \sqrt{x^2 + 9}$ has horizontal and vertical tangents. 2. **Recall the formula for the derivative:** To find tangents, we need $f'(x)$. For $f(x) = \sqrt{g(x)}$, the derivative is $$f'(x) = \frac{g'(x)}{2\sqrt{g(x)}}.$$ Here, $g(x) = x^2 + 9$. 3. **Calculate $f'(x)$:** $$g'(x) = 2x,$$ so $$f'(x) = \frac{2x}{2\sqrt{x^2 + 9}} = \frac{x}{\sqrt{x^2 + 9}}.$$ 4. **Find horizontal tangents:** Horizontal tangents occur where $f'(x) = 0$. Set $$\frac{x}{\sqrt{x^2 + 9}} = 0 \implies x = 0.$$ Calculate $f(0)$: $$f(0) = \sqrt{0^2 + 9} = 3.$$ So horizontal tangent at $y=3$ when $x=0$. 5. **Find vertical tangents:** Vertical tangents occur where $f'(x)$ is undefined or infinite. The denominator $\sqrt{x^2 + 9}$ is never zero because $x^2 + 9 \geq 9 > 0$ for all real $x$. Therefore, $f'(x)$ is defined everywhere, so no vertical tangents. 6. **Conclusion:** The function has a horizontal tangent at $y=3$ (when $x=0$) and no vertical tangents. **Answer:** d) $y=3$, no vertical tangent.