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

`a = - (1)/(2), b = - (3)/(4), c = 3`

Explanation

Given, `y = ax^(3) + bx^(2) + cx + 5` touches X-axis at `P ( - 2, 0 )` which implies that X- axis is tangent at `(-2, 0)` and the curve is also passes through `(-2, 0)`.
The curve cuts Y- axis at `(0, 5)` and gradient at this point is given 3, therefore at `(0, 5)` slope of the tangent is 3.
Now, `" " (dy)/(dx) = 3ax^(3) + 2 b x + c`
Since, X-axis is tangent at `(-2, 0)`.
`therefore " " |(dy)/(dx)|_(x = -2 ) = 0`
`rArr " " 0 = 3a (-2)^(2) + 2b (-2) + c `
`rArr " " 0 = 12 a - 4 b + c " " ` ... (i)
Again, slope of tangent at `(0, 5)` is 3.
`therefore " " |(dy)/(dx)| _("("0, 5")")= 3 `
`rArr " " 3= 3a (0)^(2) + 2b (0) + c `
`rArr " " 3= c" " `...(ii)
Since, the curve passes through `(-2, 0)`
`" " 0 = a(-2)^(3) + b (-2)^(2) + c ( - 2 ) + 5`
`rArr " " 0 = - 8a + 4b - 2c + 5 " " ` ... (iii)
From Eqs (i) and (ii),
`" " 12 a - 4b = -3 " " ` ... (iv)
From Eqs. (ii) and(iii),
`" " - 8a + 4b = 1 " " `... (v)
On adding Eqs. (iv ) and (v), we get
` 4a = -2 rArr a = - 1 //2 `
On putting `a = -1//2` in Eqs. (iv), we get
` " " 12( - 1//2) - 4b =-3 `
`rArr " " - 6 - 4b =-3`
`rArr - 3 = 4b `
`rArr " " b = - 3//4`
`therefore " " a = -1//2, b = - 3//4 and c = 3`