The point P(4,28) lies on the curve y=x^2+x+8. If Q is the point (x,x^2+x+8), find the slope of the secant line PQ for the following values of x.

Question

The point P(4,28) lies on the curve y=x^2+x+8. If Q is the point (x,x^2+x+8), 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(4,28).

Answer 1

Slope of secant = [(x² + x + 8) - 28] / (x - 4) = (x² + x - 20) / (x - 4) = (x + 5)(x - 4) / (x - 4) = x + 5

So, if x = 4.1, then slope of secant = 4.1 + 5 = 9.1, etc

Answer 2

I think that setting up a spreadsheet (e.g., using Excel) is a good way to go on this one

1st column = the 4 values of x: 4.1, 4.01, 3.99, 3.9

2nd column = calculated values of y = x² + x + 8 from adjacent values of x

3rd column = slopes calculated as (y – 28)/(x – 4)

You'll see the slope approach a certain integer as x gets closer to 4.