1. ## Secant line

The point P(4,2) lies on the curve y = x^(1/2) . If Q is the point (x,x^(1/2)) , use your calculator to find the slope of the secant line PQ (correct to 6 decimal places) for the value x=3.99.

How would I approach this?

2. Take the formula for the slope between two points.

$

m=(2-x^{\frac{1}{2}})/(4-x)

$

Then just input your value for x.

3. So it doesn't exist?

4. I think you'll find the slope does exist, and is not some huge number. 3.99 is not equal to 4, and you'll also see that the numerator is rather small as well. That's very typical of secant/tangent lines to smooth curves. You're always dealing with slopes that look like, or are close to, 0/0.

5. Originally Posted by bobsanchez
So it doesn't exist?
Well, for it to not exist would require the square root of a negative number or 0 in the denominator. It may be 0 in the denominator if you meant .99 repeating, but this would still be an infinitesmally small number.

6. Originally Posted by bobsanchez
So it doesn't exist?
you should get a value very close to 0.25