How do you find the equation of a line tangent to the function #y=(x-2)(x^2+1)# at x=-1?
Tangent Line to a Curve
the value of
and to get the point which lies on the tangent we substitute with
so the point on the line is
substitute in the following formula to get the equation
I hope this was helpful.
The tangent passes through
Since we have the gradient of the tangent and a point through which it passes we can use the point-gradient formula: