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Video Example I1) Tutorial Online Textbook EXAMPLE 1 Find an equation of the tangent line to the function y : 2x2 at the point P(1, Z). SOLUTION We will be able to nd an equation of the tangent line tas soon as we know its slope m. The difficulty is that we know only one point, P, on t, whereas we need two points to compute the slope. But observe that we can compute an approximation to m by choosing a nearby point Q(X, 2X2) on the graph and computing the slope mpQ of the secant line PQ. We choose x: 1 so that p: Q. Then :2x272 X'l "7.00 For instance, for the point 00.5, 4.5) we have mPQ=: : The tables below show the values of mp0 for several Values of x close to l. The closer Q is to P, the closer x is to 1 and, it appears from the tables, the closer mpg is to . This suggests that the slope of the tangent line t should be m = X mm: X raw 2 6 0 2 1.5 5 0.5 3 1.1 4.2 0.9 3.800 1.01 4.020 0.99 3.980 1.001 4.002 0.999 3.998 We say that the slope of the tangent line is the limit of the slopes of secant lines, and we express this symbolically by writing . 2 _ AEEVW'Q =4 and lim u :4 xal .rl Assuming that the slope of the tangent line is indeed 4, we use the point-slope form of the equation of a line (see Appendix B) to write the equation of the tangent line through (1, 2) as ya = (X71)ory= X7 . lnlS suggests tnat [ne slope of [ne tangent Ilne r snoul pe m = . X mm; X mp0 2 5 0 2 1.5 5 0.5 3 1.1 4.2 0.9 3.800 1.01 4.020 0.99 3.980 1.001 4.002 0.999 3.998 We say that the slope of the tangent line is the limit of the slopes of secant lines, and we express this symbolically by writing limmpq=4 . 2x22_ Qm and '15} .rl _4 Assuming that the slope of the tangent line is indeed 4, we use the pointislope form of the equation of a line (see Appendix B) to write the equation of the tangent line through (1, 2) as y? = (X*1)ory= xi The graphs below illustrate the limiting process that occurs in this example. A: Q approaches P along the graph, the corresponding secant lines rotate about F and approach the tangent line t. QMIPMIMM