Given that f′(2)=2 and f(2)=1 , find the equation of the tangent line to the curve y=f(x) at x=2 .

Answer 1

Since f(2) = 1, the point (2,1) is on the graph

Slope of tangent line when x = 2 is f'(2), which is given to be 2

So, using the point-slope form, an equation of the tangent line is y - 1 = 2(x - 2).

Simplify to get y = 2x - 3

Answer 2

The given information is telling you this:

Because f(2) = 1, the point on the graph of f(x) where x = 2 is (2,1).

Because f'(2) = 2, the slope of the line at (2,1) is 2. You can think of f'(x) as a "formula" for slopes of tangent lines at given x values for the function.

If you were given the problem like this:

Write the equation of the line passing through (2,1) with a slope of 2, could you do it using point-slope? Or even slope-intercept? That is essentially the essence of this problem.