Question

If `(2,6)` lies on the curve `f(x) = ax^2+bx` and `y=x+4` is a tangent to the curve at that point. Find `a` and `b`?

Answer 1

`a=-1, b=5`

Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point.

We have:

`f(x) = ax^2+bx`

We are given that the point `(2,6)` lies on the curve:

`=> f(2)=6`
`:. 4a+2b = 6`
`:. 2a+b = 3` ..... [A]

If we differentiate the parabola equation, then we have:

`f'(x) = 2ax+b`

The gradient of the tangent at `(2,6)` is given by:

`m_T = f'(2) = 4a+b`

We also know that the tangent equation is `y=x+4` and so comparing with the standard straight line equation `y=mx+c`:

`=> m_T = 1`
`:. 4a+b = 1` ..... [B]

We now solve the equations [A] and [B] simultaneous:

`[B]-[A] => 2a = -2 => a=-1`

Substitute `a=-1` into [A]:

`b-2=3 => b=5`

Hence we have:

`a=-1, b=5`