Find the Slope of the Secant Line PQ for x-values .5 to 1.001

[Destrio]

The point P(1,51) lies on the curve y=46 x2+5.

(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

 

[chaoseverlasting]

Sorry, but what is the secant line?

 

[drpizza]

Okay, I agree with your points. It may be worth looking at them in a different order:
.9, .99, .999, 1.001, 1.01, 1.1, 1.5 But that really doesn't matter in this problem.

Now, I'm wondering - for the slope of the secant line, you would probably be using the slope formula which is change in y over change in x. My question is, where on Earth did you get the point (1,13) to plug into that formula?? Did you get the (13-y)/(1-x) from another example? If so, in that example, (1,13) was a point specific to that particular example.

edit: You listed the point on the curve that eventually you're going to be wondering what the slope of the tangent is at that point. That point is (1,51), not (1,13)

edit edit: as I ran through your numbers, most actually appeared to be correct for the slope. Check those again.

I also looked at your problem another way - rather than actually plugging a y value into the slope equation, I plugged in "46x^2 + 5" for the y value (and left x as x) and simplified first. Just coincidentally, it simplified a lot, making the problem easier to calculate slopes.

 

[drpizza]

Actually, only the first couple answers have mistakes; one might be a typo. The first slope you have: 690, is off by a factor of 10.

 

[HallsofIvy]

The "secant" is a line that crosses the curve in two differnt places. If a secant crosses the line in (.5, 16.5) and (.9, 42.26) then its slope is (42.26- 16.5)/(.9- .4) = 51.52. HOW exactly did you get 690?

And, by the way, you say you are using (13- y)/(1- x), but told us that P was the point (1, 51), not (1, 13)!