1. **Problem statement:** Given the function $f(x) = 0.5x^2 - 3x - 1$, find the equation of the tangent line to the graph $G_f$ that is parallel to the given line $g$. We will solve part (a) where $g: y = -4x - 2$. 2. **Formula and rules:** The slope of the tangent line to $f$ at $x = a$ is given by the derivative $f'(a)$. For the tangent to be parallel to $g$, their slopes must be equal. The slope of $g$ is the coefficient of $x$, which is $-4$. So we solve $f'(a) = -4$. 3. **Find the derivative:** $$f(x) = 0.5x^2 - 3x - 1$$ $$f'(x) = \frac{d}{dx}(0.5x^2) - \frac{d}{dx}(3x) - \frac{d}{dx}(1) = 2 \times 0.5 x - 3 = x - 3$$ 4. **Set the slope equal to -4:** $$f'(a) = a - 3 = -4$$ $$a - 3 = -4$$ $$a = -4 + 3 = -1$$ 5. **Find the point on the graph:** Calculate $f(-1)$: $$f(-1) = 0.5(-1)^2 - 3(-1) - 1 = 0.5(1) + 3 - 1 = 0.5 + 3 - 1 = 2.5$$ So the point of tangency is $(-1, 2.5)$. 6. **Write the tangent line equation:** Using point-slope form: $$y - y_1 = m(x - x_1)$$ $$y - 2.5 = -4(x - (-1))$$ $$y - 2.5 = -4(x + 1)$$ $$y = -4x - 4 + 2.5 = -4x - 1.5$$ **Final answer:** The equation of the tangent line to $f$ parallel to $g$ is $$y = -4x - 1.5$$