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.

The curve `y=a x^3+b x^2+c x+5`touches the x-axis at `P(-2,0)`and cuts

1.	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» Correct Answer - `a = - (1)/(2), b = - (3)/(4), c = 3` 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`