1. **State the problem:** Find the equation of the tangent line to the function $h(x) = 3x^2 - 5$ at the point $(-1, 2)$. 2. **Recall the formula:** The equation of the tangent line to a function $y = f(x)$ at $x = a$ is given by: $$y = f'(a)(x - a) + f(a)$$ where $f'(a)$ is the derivative of $f(x)$ evaluated at $x = a$. 3. **Find the derivative:** Differentiate $h(x) = 3x^2 - 5$: $$h'(x) = \frac{d}{dx}(3x^2 - 5) = 6x$$ 4. **Evaluate the derivative at $x = -1$:** $$h'(-1) = 6(-1) = -6$$ 5. **Use the point and slope to write the tangent line equation:** Given point $(-1, 2)$ and slope $m = -6$, the tangent line is: $$y = -6(x - (-1)) + 2 = -6(x + 1) + 2$$ 6. **Simplify the equation:** $$y = -6x - 6 + 2 = -6x - 4$$ **Final answer:** The equation of the tangent line is: $$y = -6x - 4$$