# Slope of a Secant

• Feb 5th 2007, 01:03 PM
qbkr21
Slope of a Secant
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 $\displaystyle P(5,37)$ lies on the curve $\displaystyle y=x^2+x+7$. If $\displaystyle Q$ is the point $\displaystyle (x, x^2 + x + 7 )$, find the slope of the secant line $\displaystyle PQ$ for the following values of $\displaystyle x$:

If x=5.1, the slope of PQ is:

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

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

Thanks!
• Feb 5th 2007, 01:37 PM
Soroban
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 . . .

Quote:

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

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

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

• Feb 5th 2007, 01:39 PM
qbkr21
Soroban I am very sorry for the typo
• Feb 5th 2007, 01:44 PM
qbkr21
Ok, Now I see. So therefore and with respect to your setup the answer would be

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

Thanks!