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yanalaym [24]
3 years ago
13

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

Mathematics
1 answer:
Fynjy0 [20]3 years ago
4 0

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, (t,y(t))=(t,t^2-2t-5), where t>2. The slope of the secant line through P and Q is given by the difference quotient,

\dfrac{(t^2-2t-5)-(-5)}{t-2}=\dfrac{t^2-2t}{t-2}=\dfrac{t(t-2)}{t-2}=t

where we are allowed to simplify because t\neq2.

Then the equation of the secant line is

y-(-5)=t(x-2)\implies y=t(x-2)-5

Taking the limit as t\to2, we have

\displaystyle\lim_{t\to2}t(x-2)-5=2(x-2)-5=2x-9

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

\dfrac{\mathrm dy}{\mathrm dx}=2x-2

and at x=2, we get a slope of 2(2)-2=2, as expected.

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