The gradient function of a curve $C$ is given as \(\frac{\mathrm{d}y}{\mathrm{d}x} = 4x^3 + p

Question

The gradient function of a curve $C$ is given as $\frac{\mathrm{d}y}{\mathrm{d}x} = 4x^3 + px^2$, where $p$ is a constant. It is also known that the equation of tangent to the curve $C$ at point $A$ with coordinates $ \left(-2, \frac{16}{3} \right)$ has gradient $-4$. Show that $p = 7$.[1] Find the $x$-coordinate of stationary points on the curve and determine the nature of each stationary point.[4] Find the equation of the curve $C$.[3]

Accepted Answer

(ii)x=0, point of inflexion. x=$-\frac{7}{4}$, minimum point (iii)$y=x^4 +\frac{7}{3}x^3+8$

2024 MI H1 Q5

2024 MI H1 Q5

Solution:

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