1. The problem asks to identify the correct graph of the derivative of the function $f(x)$, where $f(x)$ is a downward-opening parabola with vertex at approximately $(1,4)$. 2. Recall that the derivative $f'(x)$ of a function $f(x)$ gives the slope of the tangent line to $f(x)$ at each point $x$. 3. For a parabola $f(x) = ax^2 + bx + c$ with $a < 0$ (downward opening), the derivative is a linear function $f'(x) = 2ax + b$. 4. Since the parabola opens downward, $a < 0$, so the derivative is a line with negative slope. 5. The vertex at $x=1$ is where the slope is zero, so $f'(1) = 0$. 6. Therefore, the derivative graph should be a straight line crossing the $x$-axis at $x=1$ with negative slope. 7. Among the given graphs, Graph 4 is a horizontal line at $y = -2$, which cannot be the derivative of a parabola. 8. Graph 2 is a parabola opening upward, which is not a linear function. 9. Graph 3 is a cubic-like curve, not linear. 10. Graph 1 is the original function, so it is not the derivative. 11. Graph 5 is unclear. 12. None of the graphs perfectly match a line with negative slope crossing $x=1$. 13. However, since the derivative of a downward parabola is a line with negative slope crossing zero at the vertex, the correct derivative graph should be a straight line crossing the $x$-axis at $x=1$. 14. If Graph 4 were a line crossing $x=1$ at zero, it would be correct, but it is a horizontal line at $y=-2$. 15. Therefore, none of the provided graphs exactly represent the derivative. Final answer: The derivative of the given parabola is a line with negative slope crossing the $x$-axis at $x=1$.