f'(x) = 2x-5
f'(1) = -3
So, the line with slope -3 through (1,-2) is
y+2 = -3(x-1)
Not sure what your four-step process is.
Use the four-step process to find the slope of the tangent line to the graph of the function at the given point and determine an equation of the tangent line.f(x)= x^2 - 5x +2 (1, -2)
2 answers
So I can just use the point-slope formula to solve equations like this?