Question

The point P(4, 23) lies on the curve y=x^{2} +x+3[/tex] . If Q is the point (x,
`x^(2)` + x + 3), find the slope of the secant line PQ for the following values of x.

If x = 4.1, the slope of PQ is: ?

and if x=4.01, the slope of PQ is: ?

and if x=3.9, the slope of PQ is: ?

and if x= 3.99, the slope of PQ is:?

Based on the above results, guess the slope of the tangent line to the curve at P(4,23) ?

There is 5 questions. Please only answer if you have all of them. It is much much appreciated.

Answer 1

The slope of the secant line PQ for different values of x are -5.9, -59.99, 4.1, and 40.99, and the slope of the tangent line to the curve at P(4, 23) is approximately -5.9.

Step-by-step explanation:

To find the slope of the secant line PQ, we need to find the difference in y-coordinates and the difference in x-coordinates between points P and Q:

If x = 4.1:

Difference in y-coordinates: (4.1)^2 + 4.1 + 3 - 23 = -0.59

Difference in x-coordinates: 4.1 - 4 = 0.1

Slope of PQ: (-0.59)/(0.1) = -5.9

If x = 4.01:

Difference in y-coordinates: (4.01)^2 + 4.01 + 3 - 23 = -0.5999

Difference in x-coordinates: 4.01 - 4 = 0.01

Slope of PQ: (-0.5999)/(0.01) = -59.99

If x = 3.9:

Difference in y-coordinates: (3.9)^2 + 3.9 + 3 - 23 = -0.41000000000000014

Difference in x-coordinates: 3.9 - 4 = -0.1

Slope of PQ: (-0.41000000000000014)/(-0.1) = 4.1

If x = 3.99:

Difference in y-coordinates: (3.99)^2 + 3.99 + 3 - 23 = -0.4099000000000002

Difference in x-coordinates: 3.99 - 4 = -0.01

Slope of PQ: (-0.4099000000000002)/(-0.01) = 40.99

Based on the results, the slope of the tangent line to the curve at P(4, 23) is approximately -5.9.