Question

The straight line with equation


`y = 11x - 28`
is a tangent to the curve

`y = {x}^(3) - a {x}^(2) + bx - 1`
at the point (3,5). Find A, the coefficient of

`{x}^(2)`
​

Answer 1

`a=3`

Explanation:

The equation of the curve is

`y=x^3-ax^2+bx-1`.

The gradient function is given by

`(dy)/(dx)=3x^2-2ax+b`

The gradient of this curve at the point (3,5) is equal to the gradient of the tangent
`y=11x-28` which is 11.

`\Rightarrow 11=3(3)^2-2a(3)+b`

`\Rightarrow 11=27-6a+b`

`\Rightarrow b-6a=-16...(1)`

The given point, (3,5) also satisfies the equation of the curve.

`\Rightarrow 3^3-a(3)^2+b(3)-1=5`

`\Rightarrow 27-9a+3b-1=5`

`\Rightarrow3b-9a=-21`

`\Rightarrow b-3a=-7...(2)`

Equation (2) minus equation (1) gives

`-3a+6a=-7+16`

`3a=9`

This implies that;

`a=3`