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.

Options

A. ` a=(1)/(2),b=-(3)/(4), c=3 `
B. ` a=-(1)/(2),b=-(3)/(4), c=3 `
C. ` a=(1)/(2),b=(3)/(4), c=3 `
D. none of these

Answer

B

Explanation

We have, ` y=ax^(3)+bx^(2)+cx +5 " "...(i) `
` rArr (dy)/(dx)=3ax^(2) + 2bx +c " "...(ii) `
Since the curve (i) touches x-axis at A(-2,0). Therefore, point A lies on (i) and `((dy)/(dx))_(A)=0.`
` therefore -8a+4b-2c+5=0 " "...(iii) `
and, ` 12a-4b+c=0 " "...(iv) `
The curve (i) cuts y-axis at B(0,5). It is given that
` ((dy)/(dx))_(B)=3 rArr c=3 `
Putting c = 3 in (iii) and (iv) and solving them, we get
` a=-(1)/(2) " and " b=-(3)/(4) `