# Find the two points on the graph f(x)=x^3+6x^2-18x-100 where the gradient is 18. Write the equations of tangents at these points.

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### 1 Answer

`f(x) = x^3+6x^2-18x-100`

The gradient can be found by using the first derivative. Therefore.

`f'(x) = 3x^2+12x-18 = 18`

This gives,

`3x^2+12x-36 = 0`

`x^2+4x-12 = 0`

`(x+6)(x-2) = 0`

`x = -6 or x = 2`

Therefore there are two points at which the gradient is 18.

**1) At** `x = -6`

`f(-6) = (-6)^3+6(-6)^2-18(-6)-100`

`f(-6) = -216+216+108-100`

`f(-6) = 8`

The tangent through `x = -6` can be written as,

`y = 18x + c`

We know at `x = -6` , `y = 8` , then,

`8 = 18(-6)+c`

`c = 108+8`

`c = 116`

Therefore the tangent through `x =-6` , is

`y = 18x +116`

**2) At** `x = 2`

`f(2) = 2^3+6xx2^2-18xx2-100`

`f(2) = 8+24-36-100`

`f(2) = -104`

The tangent through `x = 2` can be written as,

`y = 18x + k`

We know at `x = 2, y = -104,` then,

`-104 = 18xx2+k`

`k = -104-36`

`k = -140`

Therefore the tangent through `x =2` is,

`y = 18x -140`

**Therefore the two tangents are `y = 18x+116` and `y =18x-140` **