Using f(x)=x^2-2x+3, we can determine that f'(x)-2x-2. This means that the​ point-slope equation of the tangent line to the graph of f(x)=x^2-2x+3 at (6,27) is

Question

y - _________ = _________(x - _________).

Answer 1

The derivative is the slope of the tangent line at any value of "x" so if the derivative is 2x - 2 then at x = 6, the derivative is 2(6) - 2 = 10. So the slope is 10. A point on the line is the point of tangency (6, 27) so now you can use the point-slope form of a line to get the equation of the tangent line.

The point-slope form is y - y₁ = m(x - x₁) where "m" is the slope and the point is (x₁, y₁)

So the equation (where m = 10, x₁ = 6, and y₁ = 27) is:

y - 27 = 10(x - 6)