Parabola Tangent A0B571

1. **State the problem:** We are given the function $f(x) = x^2 - 14x + 49$ and need to sketch its graph and the tangent line at a specific point (from part (b), which we assume is a point on the curve). 2. **Rewrite the function:** Notice that $f(x) = x^2 - 14x + 49$ can be rewritten by completing the square: $$f(x) = (x^2 - 14x + 49) = (x - 7)^2$$ This is a perfect square, so the graph is a parabola with vertex at $(7,0)$. 3. **Find the derivative:** The slope of the tangent line at any point $x$ is given by the derivative: $$f'(x) = \frac{d}{dx} (x - 7)^2 = 2(x - 7)$$ 4. **Find the point from part (b):** Since the user did not specify the point, we assume the point is at $x=7$ (vertex) or another point. For demonstration, let's use $x=7$. 5. **Calculate the slope at $x=7$:** $$f'(7) = 2(7 - 7) = 0$$ 6. **Find the point coordinates:** $$f(7) = (7 - 7)^2 = 0$$ So the point is $(7,0)$. 7. **Equation of the tangent line:** Using point-slope form: $$y - y_1 = m(x - x_1)$$ where $m = 0$ and $(x_1,y_1) = (7,0)$, $$y - 0 = 0(x - 7)$$ $$y = 0$$ 8. **Summary:** The graph is a parabola $y = (x-7)^2$ with vertex at $(7,0)$. The tangent line at $x=7$ is the horizontal line $y=0$. **Final answer:** - Function graph: $y = (x-7)^2$ - Tangent line at $(7,0)$: $y=0$