What is the tangent line of y=x^3-3x+1 at (2,3)

Remember
the equation of a line that passes through the point (x1,y1) and has a slope of m is
y-y1=m(x-x1)

so
find the slope at (2,3)
take the deritivive
dy/dx=3x²-3
(2,3)
at x=2
dy/dx=3(2)²-3
dy/dx=3(4)-3
dy/dx=12-3
dy/dx=9

slope is 9 at that point

point given is (2,3)
y-3=9(x-2)
y-3=9x-18
y=9x-15 is the answer

The linear equation is: y=ax+b