The point P(9, 1) lies on the curve y = ?x

The point P(9, 1) lies on the curve y = ?x - 8. (a) If Q is the point (x, ?x - 8), use your calculator to find the slope of the secant line PQ (correct to six decimal places) for the following values of x. (i) 8.5 (ii) 8.9 (iii) 8.99 (iv) 8.999 (v) 9.5 (vi) 9.1 (vii) 9.01 (viii) 9.001 (b) Using the results of part (a), guess the value of the slope of the tangent line to the curve at P(9, 1). (c) Using the slope from part (b), find an equation of the tangent line to the curve at P(9, 1). y =

Transcript

00:01 Here, we're given y equals the square root of x minus 8.

00:05 And we're told that if x is equal to 9, the y value is equal to 1, which makes sense.

00:11 We could plug that in and convert it ourselves.

00:14 Well, we want to use our calculator to find the slope of the sinket line p to q.

00:19 Let's go ahead and kind of give a little bit of a graph.

00:21 The square root of x minus 8 means that we're going to shift it to the right 8 from a square root.

00:27 So it looks something like this.

00:28 So we have to plug in numbers that are bigger than 8.

00:34 So let's find now we're looking at the point 9.

00:39 So that's say right here.

00:42 So we're going to look at different points, say 8 .5 is say here, and we want to find the slope here.

00:48 Well, that's going to be y2 minus y1 over x2 minus x1.

00:55 So let's say that we have the point 8 .5.

01:01 If x is equal to 8 .5, we have y2, which is the square root of 8 .5 minus y1, which is 1.

01:13 That comes from the original point.

01:15 We're just finding the slope between these two values, divided by x2, which is 8 .5, minus x1, which is 9.

01:25 If i plug that into my calculator, and i'm just going to use the calculator because it says i can, i get that 8 .5, i get that 8 .5 .5, minus x1, which is 9.

01:33 If i plug that into my calculator, and i'm just going to use the calculator, i get that 8...

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