The diagram shows the graph of $y=\mathrm{f}'\left( x \right)$, which passes through the origin an

Question

The diagram shows the graph of $y=\mathrm{f}'\left( x \right)$, which passes through the origin and has turning points at $(2,5)$ and $(6,0)$. The graph of $y={{\mathrm{f}}'}\left( x \right)$ has a vertical asymptote $x=-1$ and a horizontal asymptote $y=5$. Sketch the graph of ${y = \mathrm{f}''(x)}$, stating the equation of asymptote(s) and the coordinates of any intersection(s) with the axes where possible.[3] Write down the $x$-coordinate(s) of the stationary point(s) of $y=\mathrm{f}\left( x \right)$ and determine their nature.[2] If the tangent to the graph of $y=\mathrm{f}\left( x \right)$ at $x=2$ passes through the origin, find the equation of the normal to the graph of $y=\mathrm{f}\left( x \right)$ at $x=2$.[2]

Accepted Answer

(i) $x=-1$ is a vertical asymptote, $y=0$ is a horizontal asymptote, and the $x$-intercepts are $(2,0)$ and $(6,0)$ (ii) $x=0$, minimum point; $x=6$, stationary point of inflexion (iii) $\displaystyle y=-\frac{1}{5}x+\frac{52}{5}$

2024 ACJC Promo Q4

2024 ACJC Promo Q4

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