Find the slope of the curve y=x^2-2x-5 at the point p(2,5) by finding the limit of secant slopes through point p

The point (2, 5) is not on the curve; probably you meant to say (2, -5)?

Consider an arbitrary point Q on the curve to the right of P, , where . The slope of the secant line through P and Q is given by the difference quotient,

where we are allowed to simplify because .

Then the equation of the secant line is

Taking the limit as , we have

so the slope of the line tangent to the curve at P as slope 2.

- - -

We can verify this with differentiation. Taking the derivative, we get

and at , we get a slope of , as expected.

mr. graef told us it is 13 pints

step-by-step explanation: