1. **State the problem:** Find the equation of the tangent line to the graph of $f(x)=\sqrt{2x^2+1}$ at $x=-1$. 2. **Recall the formula for the tangent line:** The equation of the tangent line to $y=f(x)$ at $x=a$ is given by $$y = f(a) + f'(a)(x - a)$$ where $f'(a)$ is the derivative of $f$ evaluated at $x=a$. 3. **Find $f'(x)$:** Given $f(x) = \sqrt{2x^2 + 1} = (2x^2 + 1)^{1/2}$, using the chain rule, $$f'(x) = \frac{1}{2}(2x^2 + 1)^{-1/2} \cdot 4x = \frac{2x}{\sqrt{2x^2 + 1}}$$ 4. **Evaluate $f(-1)$:** $$f(-1) = \sqrt{2(-1)^2 + 1} = \sqrt{2 + 1} = \sqrt{3}$$ 5. **Evaluate $f'(-1)$:** $$f'(-1) = \frac{2(-1)}{\sqrt{2(-1)^2 + 1}} = \frac{-2}{\sqrt{3}}$$ 6. **Write the tangent line equation:** $$y = f(-1) + f'(-1)(x - (-1)) = \sqrt{3} - \frac{2}{\sqrt{3}}(x + 1)$$ 7. **Simplify the equation:** $$y = \sqrt{3} - \frac{2}{\sqrt{3}}x - \frac{2}{\sqrt{3}}$$ Combine constants: $$\sqrt{3} - \frac{2}{\sqrt{3}} = \frac{3}{\sqrt{3}} - \frac{2}{\sqrt{3}} = \frac{1}{\sqrt{3}}$$ So, $$y = - \frac{2}{\sqrt{3}}x + \frac{1}{\sqrt{3}}$$ **Final answer:** $$\boxed{y = - \frac{2}{\sqrt{3}}x + \frac{1}{\sqrt{3}}}$$