Question

If g(x) is differentiable so that g(-4)=0 and g'(-4)=3, what would the equation of a tangent line to y=g(x) at x=4 be?
I listed the steps I took but it was wrong, can someone tell me how to figure this one out? Thanks!

The steps I took:
g(x)=g(-4)+g'(4)*(x-4)
g(x)=0-4(x-4)
g(x)=16-4x

Answer 1

Is the question
a tangent line to y=g(x) at x= 4 be? or x= -4 ??

hopefully at x= -4 because that is where we know the slope

when x=-4 g(x)= 0, means (-4,0) is a point on the curve.
the slope at that point is g'(-4) = 3

you have a point and a slope. use
y - y0 = m(x-x0)
y - 0 = 3(x - -4)
y = 3x +4

Answer 2

@phi it is x= -4 my bad! Thanks, I'll look this over in a bit.