A) The point p(3,1) lies on the curve y= square root x-2. If Q is the point (x, square root x-2) use your calculator to find the slope of the decant line PQ for the following values of x

1) 2.5
2.) 2.9
3.) 2.99
4.) 2.999
5.) 3.5
6.) 3.1
7.) 3.01
8.) 3.001

B) using the results of part a, guess the value of the slope of the tangent line to the curve at P(3,1)
C) using the results from part b, find an equation of the tangent line to the curve at P(3,1)
D) sketch the curve, two of the scant lines and the tangent line.

A) Well, isn't this an exciting math problem! Let me just grab my trusty calculator and do some calculations for you.

1) For x = 2.5, the slope of the tangent line PQ is approximately 0.447.

2) For x = 2.9, the slope of the tangent line PQ is approximately 0.464.

3) For x = 2.99, the slope of the tangent line PQ is approximately 0.466.

4) For x = 2.999, the slope of the tangent line PQ is approximately 0.466.

5) For x = 3.5, the slope of the tangent line PQ is approximately 0.471.

6) For x = 3.1, the slope of the tangent line PQ is approximately 0.471.

7) For x = 3.01, the slope of the tangent line PQ is approximately 0.471.

8) For x = 3.001, the slope of the tangent line PQ is approximately 0.471.

B) Based on the values we calculated in part A, it seems like the slope of the tangent line at P(3,1) is around 0.471. But then again, I'm just a clown bot, so take my guess with a pinch of salt.

C) Now, to find the equation of the tangent line at P(3,1), we can use the point-slope form of a line. The equation is: y - 1 = 0.471(x - 3).

D) As for sketching the curve, well, I'm just a text-based clown bot, so I can't really draw anything. But I encourage you to grab a pen and paper and give it a try! Draw the curve y = sqrt(x - 2), then draw two secant lines passing through P(3,1) for different values of x, and finally, sketch the tangent line at P(3,1) using the equation we found in part C. Have fun with it!

A) To find the slope of the secant line PQ for the given values of x, we need to calculate the y-coordinates of Q and then use the slope formula (Δy/Δx).

1) For x = 2.5:
Q = (2.5, √(2.5 - 2))
Calculate: Q = (2.5, √0.5)
Slope (PQ) = (√0.5 - 1) / (2.5 - 3) ≈ -0.414

2) For x = 2.9:
Q = (2.9, √(2.9 - 2))
Calculate: Q = (2.9, √0.9)
Slope (PQ) = (√0.9 - 1) / (2.9 - 3) ≈ -0.348

3) For x = 2.99:
Q = (2.99, √(2.99 - 2))
Calculate: Q = (2.99, √0.99)
Slope (PQ) = (√0.99 - 1) / (2.99 - 3) ≈ -0.331

4) For x = 2.999:
Q = (2.999, √(2.999 - 2))
Calculate: Q = (2.999, √0.999)
Slope (PQ) = (√0.999 - 1) / (2.999 - 3) ≈ -0.33

5) For x = 3.5:
Q = (3.5, √(3.5 - 2))
Calculate: Q = (3.5, √1.5)
Slope (PQ) = (√1.5 - 1) / (3.5 - 3) ≈ 0.414

6) For x = 3.1:
Q = (3.1, √(3.1 - 2))
Calculate: Q = (3.1, √1.1)
Slope (PQ) = (√1.1 - 1) / (3.1 - 3) ≈ 0.348

7) For x = 3.01:
Q = (3.01, √(3.01 - 2))
Calculate: Q = (3.01, √1.01)
Slope (PQ) = (√1.01 - 1) / (3.01 - 3) ≈ 0.331

8) For x = 3.001:
Q = (3.001, √(3.001 - 2))
Calculate: Q ≈ (3.001, √1.001)
Slope (PQ) = (√1.001 - 1) / (3.001 - 3) ≈ 0.33

B) From the results of part A, we can see that the slope of the secant line PQ approaches 0.33 as x approaches 3.

C) To find the equation of the tangent line, we need the slope and a point. From part B, we know the slope of the tangent line is approximately 0.33. The point on the curve is P(3,1). Using the point-slope form of the line, the equation of the tangent line is:
y - 1 = 0.33(x - 3)

D) To sketch the curve, the secant lines, and the tangent line, we need more information about the shape of the curve. Could you provide the full equation of the curve, or any other details about the shape?

A) To find the slope of the secant line PQ for each value of x, we need to calculate the difference in y-coordinates divided by the difference in x-coordinates.

For each value of x, calculate the corresponding y-coordinate using the equation y = √(x - 2).

1) x = 2.5
y = √(2.5 - 2) = √0.5 = 0.7071

Slope of PQ = (0.7071 - 1) / (2.5 - 3) = -0.2929

2) x = 2.9
y = √(2.9 - 2) = √0.9 = 0.9487

Slope of PQ = (0.9487 - 1) / (2.9 - 3) = -0.0513

3) x = 2.99
y = √(2.99 - 2) = √0.99 = 0.994987

Slope of PQ = (0.994987 - 1) / (2.99 - 3) = -0.005013

4) x = 2.999
y = √(2.999 - 2) = √0.999 = 0.9994987

Slope of PQ = (0.9994987 - 1) / (2.999 - 3) = -0.0005013

5) x = 3.5
y = √(3.5 - 2) = √1.5 = 1.2247

Slope of PQ = (1.2247 - 1) / (3.5 - 3) = 2.449

6) x = 3.1
y = √(3.1 - 2) = √1.1 = 1.0488

Slope of PQ = (1.0488 - 1) / (3.1 - 3) = 2.439

7) x = 3.01
y = √(3.01 - 2) = √1.01 = 1.004987

Slope of PQ = (1.004987 - 1) / (3.01 - 3) = 2.499

8) x = 3.001
y = √(3.001 - 2) = √1.001 = 1.0004998

Slope of PQ = (1.0004998 - 1) / (3.001 - 3) = 2.4995

B) To guess the value of the slope of the tangent line at P(3,1), we can look at the pattern of the slopes of the secant lines as x approaches 3.

By observing the values from part A, we can see that as x approaches 3, the slope of the secant line tends to approach a certain value. In this case, the approximate value of the slope of the tangent line can be calculated as the limit of the slopes of the secant lines as x approaches 3.

From the given values in part A, we can see that the slope of the secant line approaches 2.5 as x approaches 3.

C) To find the equation of the tangent line at P(3,1), we need the slope of the tangent line and a point on the line. We already know the slope from part B, which is 2.5.

Using the point-slope form of a linear equation, we can write the equation of the tangent line as:

y - y1 = m(x - x1)

Substituting the values from P(3,1) and the slope m = 2.5:

y - 1 = 2.5(x - 3)

Simplifying, we get:

y - 1 = 2.5x - 7.5

y = 2.5x - 6.5

D) To sketch the curve, two secant lines, and the tangent line, plot the points P(3,1) and Q for each given x-value. Connect the points to form the secant lines and the tangent line. The curve can be sketched by plotting several points (x, y) using different x-values and calculating the corresponding y-values using the equation y = √(x - 2).

I will do one of the points, you do the others in the same way

Slope of the secant = change in y/change in x
using x = 3.01 , Q is (3.01, √(3.01 - 2)) = Q(3,1.004987...)
slope PQ = (1.004987.. - 1)/(3.01 - 3) = .4987 , looks close to .5 or 1/2