Question

Find the equation of the tangent line the the graph of f at the given point: f(v) = (v-4)(v^2-5), at point (-3,-28).

Answer 1

so get the derivative

Answer 2

@julian25 -- I've done that all ready. \[f'(v) = [1*(v^2-5)]+[(v-4)*2v]\]\[= v^2-5+2v^2-8v\]\[=3v^2-8v-5\]

Answer 3

once you have the derivative plug in the values given at the beginning of the problem

Answer 4

Does that make sense? Let me know.

Answer 5

Yes, it makes sense; so, I would do the following: \[f'(28)=3(-3)^2-8(-3)-5\]\[=-8\] Or did I make a mistake?

Answer 6

I believe that is correct...I don't recall the exact formula but try that answer