Question

Not EASY: The point P(25 , 7 ) lies on the curve y =sqrt(x)+ 2. Let Q be the point (x, sqrt(x)+ 2 ).

a.) Find the slope of the secant line PQ for the following values of x. (Answers here should be correct to at least 6 places after the decimal point.)

If x= 25.1, the slope of PQ is:
If x= 25.01, the slope of PQ is:
If x= 24.9, the slope of PQ is:
If x= 24.99, the slope of PQ is:

b.) Based on the above results, estimate the slope of the tangent line to the curve at P(25 , 7 ).

Answer 1

If x = 25.1, y = 7.01
Slope of PQ = \[\frac{25.1-25}{7.01-7}=\frac{.1}{.01}=10\]

Answer 2

Find the other slopes similarly and predict the slope of the tangent based upon whatever the slopes of the secants are close to.

Answer 3

Hey, thanks, but why 7.01?

Answer 4

Because if you replace x with 25.1, that is what you get for y. You could carry more significant digits if you wish but you are really not entitled to more.

Answer 5

We have:
\[y=\sqrt{x}+2\]
We also have two points. P, and Q: \(P(25,7)\) and \(Q(x,\sqrt{x}+2)\)
The slope, \(m\) of P and Q is:
\[m=\frac{y_2-y_1}{x_2-x_1}=\frac{(\sqrt{x}+2)-7}{x-25}=\frac{\sqrt{x}-5}{x-25}\] as it gets near 25. So we can do like so:
\[f(x)=\frac{\sqrt{x}-5}{x-25}\]
And calculate f(25.1), f(25.01), f(24.9), and f(24.99)

Answer 6

use a spread sheet