The point P(1,51) lies on the curve y=46 x2+5.(adsbygoogle = window.adsbygoogle || []).push({});

(a) If Q is the point (x,46 x2+5), use your calculator to find the slope of the secant line PQ (correct to six decimal places) for the following values of x.

.5, .9, .99, .999, 1.5, 1.1, 1.01, 1.001

I plugged all of the values of x into point Q, getting values of

(.5, 16.5)

(.9, 42.26)

(.99, 50.0846)

(.999, 50.908046)

(1.5, 108.5)

(1.1, 60.66)

(1.01, 51.9246)

(1.001, 51.092046)

I then did (13- y)/(1-x) for each of the points getting:

690

87.5

91.54

91.959

115

96.6

92.46

92.046

Somewhere I have gone wrong

Any help would be much appreciated

Thanks

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# Homework Help: Tangent Problem

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