1. **Problem Statement:** Find the horizontal and vertical tangents of the function $f(x) = \sqrt{x^2 + 9}$. 2. **Formula and Rules:** The derivative $f'(x)$ gives the slope of the tangent line. Horizontal tangents occur where $f'(x) = 0$. Vertical tangents occur where $f'(x)$ is undefined. 3. **Find the derivative:** $$f(x) = \sqrt{x^2 + 9} = (x^2 + 9)^{1/2}$$ Using the chain rule: $$f'(x) = \frac{1}{2}(x^2 + 9)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + 9}}$$ 4. **Find horizontal tangents:** Set $f'(x) = 0$: $$\frac{x}{\sqrt{x^2 + 9}} = 0 \implies x = 0$$ Calculate $f(0)$: $$f(0) = \sqrt{0 + 9} = 3$$ So horizontal tangent at $y=3$ when $x=0$. 5. **Find vertical tangents:** Check where $f'(x)$ is undefined. The denominator $\sqrt{x^2 + 9}$ is never zero since $x^2 + 9 \geq 9 > 0$. So no vertical tangents. 6. **Conclusion:** Horizontal tangent at $y=3$ when $x=0$. No vertical tangents. 7. **Answer:** Option (d) $y=3$, no vertical tangent.