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.

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

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

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b

answer: translate the circles so they share a common center point and dilate circle y by a scale factor of 3.