If f'(-3)=2 and f(-3)=5, write the equation of the normal line to y=f(x) at x=-3

Answer 1

The tangent at a point on a function will have the slope f'(a) and pass through the point (a, f(a))

Point-slope form:

y - f(a) = f'(a)(x-a)

Here a = -3 f(a) = 5 and f'(a) = 2

Please consider a tutor. Take care.