Question

How do you find the equation of the tangent to the curve with equation `y = x^(3) + 2x^(2) - 3x + 2` at the point where x = 1?

Answer 1

Tangent has equation: `y=4x-2`

Explanation:

First we calculate `f(1)` to find the point at which tangent and the curve meet.

`f(1)=1+2-3+2=2`

So we are looking for a tangent in `(1,2)`.

The tangent line has equation of: `y=ax+b`, where `a=f'(1)`, so we have to calculate `f'(1)`.

`f'(x)=3x^2+4x-3`
`f'(1)=3+4-3=4`

So the tangent line has equation `y=4x+b`, now we are looking for such `b`, for which the tangent line passes through `(1,2)`

`2=4+b`

`b=-2`

Finally the tangent line has equation of `y=4x-2`

graph{(y-x^3-2x^2+3x-2)(y-4x+2)=0 [-17.1, 11.37, -3.28, 10.96]}