Question

How do you find the equation of tangent line to the curve `f(x)=2x^2` at x=-1?

Answer 1

Equation of tangent is `4x+y+2=0`

Explanation:

At `x=1` ,`f(x)=2xx(-1)^2=2` hence tangent passes through `(-1,2)`

Slope of the line is given by value of derivative and as

`f'(x)=4x`, slope of tangent at `(-1,2)` will be `4xx(-1)=-4`

Hence equation of tangent is

`(y-2)=-4(x+1)` or

`4x+y+2=0`