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.

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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.

4 0

3 years ago

What is the answer to: x-2 ___= 3 5

Alika [10]

(x - 2)/5 = 3

Multiply 5 to both sides:

x - 2 = 15

Add 2 to both sides:

x = 17

6 0

4 years ago

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