Tangent line help please?

Question

Let f be a differentiable function such that f(3) = 2 and f '(3) = 10. If the tangent line to the graph of f at x = 3 is used to find an approximation to a zero of f, that approximation would be what?

OK, so I found the tangent line is y = 10x - 32. What do I do from here?

Thanks!

Answer 1

The tangent line has gradient 10 so is of the form

y=10x+c

When x=3, y=2

=>c=-28

so the tangent line has equation

y=10x-28

For y=0

=>x=2.8

so x=2.8 is an approximation to a zero of f.