1. **Problem:** Given a function $f$ with $f(3) = -5$ and slope at any point $(x,y)$ on the graph given by $\frac{2x^2}{y}$, find the equation of the tangent line at $x=3$ and use it to approximate $f(2.9)$. 2. **Formula:** The tangent line at $x=a$ is given by $$y = f(a) + f'(a)(x - a)$$ where $f'(a)$ is the slope of the tangent line at $x=a$. 3. **Find $f'(3)$:** Since slope $= \frac{2x^2}{y}$, at $x=3$, $y=f(3)=-5$, so $$f'(3) = \frac{2 \cdot 3^2}{-5} = \frac{18}{-5} = -\frac{18}{5} = -3.6$$ 4. **Equation of tangent line at $x=3$:** $$y = f(3) + f'(3)(x - 3) = -5 - 3.6(x - 3)$$ 5. **Approximate $f(2.9)$:** Substitute $x=2.9$: $$f(2.9) \approx -5 - 3.6(2.9 - 3) = -5 - 3.6(-0.1) = -5 + 0.36 = -4.64$$ **Final answer:** The tangent line is $$y = -5 - 3.6(x - 3)$$ and the approximation is $$f(2.9) \approx -4.64$$.