Use the equation m Subscript PQ Baseline equals StartFraction f left parenthesis x 1 plus h right parenthesis minus f left paren
thesis x 1 right parenthesis Over h EndFraction mPQ= fx1+hâfx1 h to calculate the slope of a line tangent to the curve of the function y equals f left parenthesis x right parenthesis equals 2 x squared y=f(x)=2x2 at the point Upper P left parenthesis x 1 comma y 1 right parenthesis equals Upper P left parenthesis 3 comma 18 right parenthesis Px1,y1=P(3,18).
The attached graph shows a numerical estimate of the slope (f'(3)=12) and the line with that slope through the point P(3, 18). The line certainly appears to be tangent to the curve at that point.