Slope of a Secant

**qbkr21** · February 5th 2007, 01:03 PM

There are several of these problems-all of which are very closely related. If someone could please get me started it would help me out alot. The problems are as follows:...

The point $P(5,37)$ lies on the curve $y=x^2+x+7$ . If $Q$ is the point $(x, x^2 + x + 7 )$ , find the slope of the secant line $PQ$ for the following values of $x$ :

If x=5.1, the slope of PQ is:

Ohh Yea one other thing the formula for the slope of secant line is:

$m_{PQ} = \frac { y_{Q} - y_{P}}{x_{Q} - x_{P}}$

Thanks!

**Soroban** · February 5th 2007, 01:37 PM

Hello, qbkr21!

This problem is straight-forward ... and you know the Slope Formula.
So exactly where is your difficulty?

And I assume there is a typo in your problem . . .

The point $P$ (5, 37) lies on the curve $y\:=\:x^2+x+7$ .
Find the slope of the secant line $PQ$ for $x = 5.1$

When $x = 5.1\!:\;\;y \:=\:5.1^2 + 5.1 + 7 \:=\:38.11$
. . Hence, point $Q$ is $(5,1,\,38.11)$

Can you find the slope of $PQ$ now?
. . $P(5,37),\;Q(5.1,\,38.11)$

**qbkr21** · February 5th 2007, 01:39 PM

Soroban I am very sorry for the typo

**qbkr21** · February 5th 2007, 01:44 PM

Ok, Now I see. So therefore and with respect to your setup the answer would be

$\frac{38.11-37}{5.1-5}$ $=$ $11.1$

Thanks!

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