The curve $y=a x^3+b x^2+c x+5$ touches X-axis at $P(-2,0)$ and cuts $\mathrm{Y}$-axis at a point $\mathrm{Q}$, where its gradient is 3 , then

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Aicharya · Accepted Answer

The curve $y=a x^3+b x^2+c x+5$ touches $X$-axis at $P(-2,0)$ $	herefore 0=\mathrm{a}(-2)^3+\mathrm{b}(-2)^2+\mathrm{c}(-2)+5$ $\frac{d y}{d x}=3 a x^2+2 b x+c$ and at point $Q$ on $Y$ axis, we have $\frac{d y}{d x}=3$. Let $Q \equiv(0, k)$ The equation (1) becomes $8 a-4 b+6=5$ i.e. $8 a-4 b=-1$ $\Rightarrow 2 \mathrm{a}-\mathrm{b}=\frac{-1}{4}$ At $\mathrm{P}(-2,0)$ $\frac{\mathrm{dy}}{\mathrm{dx}}=0$ From (1), (2) \& (3) $\mathrm{a}=\frac{-1}{2}, \mathrm{~b}=\frac{-3}{4}, \mathrm{c}=3$

The curve $y=a x^3+b x^2+c x+5$ touches X-axis at $P(-2,0)$ and cuts $\mathrm{Y}$-axis at a point…

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