A curve has equation $$y=ax^3+bx^2+cx+5.$$ Its gradients at $x=0$, $x=1$, and $x=2$ are $0$,

Question

A curve has equation $$y=ax^3+bx^2+cx+5.$$ Its gradients at $x=0$, $x=1$, and $x=2$ are $0$, $1$, and $8$, respectively. Find $a$, $b$, and $c$. Hence find all points on the curve where the tangent is horizontal.

Accepted Answer

(a) $a=1,\ b=-1,\ c=0$; (b) Horizontal tangents at $(0,5)$ and $\left(\frac{2}{3}, \frac{131}{27}\right)$.

Differentiation

Differentiation

Solution:

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