Question

How do you find the equation of a line tangent to the function `y=(x-2)(x^2+1)` at x=-1?

Answer 1

`y=8x+2`

Explanation:

Tangent Line to a Curve
simplify
`y=x^3-2x^2+x-2`
differentiate
`y'=3x^2-4x+1`
substitute for `x=-1`
`y'=8`
the value of `y'` is the slope of the tangent line at `x=-1`
and to get the point which lies on the tangent we substitute with `x=-1` in the function
`y=-6`
so the point on the line is`(-1,-6)` and its slope is 8
substitute in the following formula to get the equation

`y-y_1=m(x-x_1)`

`y+6=8(x+1)`

`y=8x+2`

I hope this was helpful.

Answer 2

`y = 8x + 2`

Explanation:

At `x = -1`:

`y = (-1-2)((-1)^2 + 1)`

`y = -3 xx 2`

`y = -6`

The tangent passes through `(-1,-6)`

The gradient:

`y = (x-2)(x^2 + 1)`

`y = x^3 - 2x^2 + x - 2`

`y' = 3x^2 - 4x + 1`

At `x = -1`:

`m = 3(-1)^2 - 4(-1) + 1`

`m = 3 + 4 + 1`

`m = 8`

Since we have the gradient of the tangent and a point through which it passes we can use the point-gradient formula:

`y - y_1 = m(x - x_1)`

`y +6 = 8(x +1 )`

`y = 8x + 8 - 6`

`y = 8x + 2`