The curve `y=a x^3+b x^2+c x+5` touches the x-axis at `P(-2,\ 0)` and cuts the y-axis at the point `Q` where its gradient is 3. Find the equation of the curve completely.

Answer

Equation of curve is

`y=frac{-x^{3}}{2}-frac{3}{4} x^{2}+3 x+5`

Explanation

ATQ, it is given,

`y=a x^{3}+b x^{2}+c x+5`


As `(-z, 0)` also lies on curve `y=a x^{3}+b x^{2}+c x+5 d`


`\Rightarrow 0 = - 8a + 4b - 2c + 5`


Also at `(-2,0)` curve touches `x`-axis i.e.

`\frac{d y}{d x}` at `(-2,0)` is 0


`\therefore \frac{d y}{d x}=3 a x^{2}+2 b x+c `


`\frac{d y}{d x}(-2,0)=0=12 a=-4 b+c .`


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