How do you find the tangent line of f(x)=3-2x at x=-1?

Explanation:
At x=-1, f(-1)=3-2(-1)=5
So the tangent touches the function at the point (-1,5).
The gradient of the tangent is the derivative of the function.
${\displaystyle \therefore }$ f'(x)=-2 and in particular f'(-1)=-2.
The tangent is a straight line so has equation y=mx+c.
We may substitute the point (-1,5) in to obtain
${\displaystyle 5=(-2)(-1)+c\Rightarrow c=3.}$
Hence the equation of the required tangent line to the function at the given point is
y=-2x+3.