1. **Problem:** Given a function $f$ with $f(3) = -5$ and the slope of the tangent line at any point $(x,y)$ on the graph is 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 and rules:** The equation of the tangent line to $f$ 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 the slope at $x=3$:** We know $f(3) = -5$, so the point is $(3, -5)$. The slope at this point is: $$f'(3) = \frac{2 \cdot 3^2}{-5} = \frac{18}{-5} = -\frac{18}{5}$$ 4. **Write the tangent line equation:** Using the point-slope form: $$y - (-5) = -\frac{18}{5}(x - 3)$$ Simplify: $$y + 5 = -\frac{18}{5}(x - 3)$$ 5. **Approximate $f(2.9)$:** Substitute $x=2.9$ into the tangent line: $$y = -5 - \frac{18}{5}(2.9 - 3) = -5 - \frac{18}{5}(-0.1) = -5 + \frac{18}{5} \times 0.1 = -5 + \frac{18}{50} = -5 + 0.36 = -4.64$$ **Final answer:** The tangent line equation at $x=3$ is: $$y + 5 = -\frac{18}{5}(x - 3)$$ The approximation for $f(2.9)$ is: $$f(2.9) \approx -4.64$$