Given f(x)=(x-3)^2-4 with one tangent line with a slope=2, how do you find the point of tangency?

Answer 1

You look for the value(s) of x where the derivative = 2.

f'(x) = 2(x - 3)
2 = 2x - 6
8 = 2x
x = 4

Plug back in the original equation:

f(4) = (4-3)^2 - 4 = 1 - 4= -3

So the point of tangency is (4, -3)

Answer 2

f(x)=(x-3)^2-4=x^2-6x+5
f'(x)=2x-6=2
2x=8
x=4
f(4)=(4-3)^2-4=-3

point of tangency is (4,-3)