1. Problem: Find the equations of the tangent lines to the function $f(x) = -x^2 + 4x - 14$ at the points $(0, f(0))$ and $(7, f(7))$. 2. Formula: The slope of the tangent line at a point $x=a$ is given by the derivative $f'(a)$. The tangent line equation is: $$y = f'(a)(x - a) + f(a)$$ 3. Find $f(0)$ and $f(7)$: $$f(0) = -(0)^2 + 4\cdot0 - 14 = -14$$ $$f(7) = -(7)^2 + 4\cdot7 - 14 = -49 + 28 - 14 = -35$$ 4. Find the derivative $f'(x)$: $$f'(x) = \frac{d}{dx}(-x^2 + 4x - 14) = -2x + 4$$ 5. Calculate slopes at $x=0$ and $x=7$: $$f'(0) = -2\cdot0 + 4 = 4$$ $$f'(7) = -2\cdot7 + 4 = -14 + 4 = -10$$ 6. Write tangent line equations: At $(0, -14)$: $$y = 4(x - 0) - 14 = 4x - 14$$ At $(7, -35)$: $$y = -10(x - 7) - 35 = -10x + 70 - 35 = -10x + 35$$ Final answers: - Tangent at $(0, -14)$: $y = 4x - 14$ - Tangent at $(7, -35)$: $y = -10x + 35$