# Thread: How to find the slopes of:

1. ## How to find the slopes of:

y = (3x + 1)^2

and

y = (2x - 5) / (3x + 2)^2

2. Originally Posted by marie7
y = (3x + 1)^2

and

y = (2x - 5) / (3x + 2)^2
Not enough info marie. Do you have any relevant points. Or you are trying to find a derivative, or what...

Let me know, and I'll jump on it.

3. omg oops I forgot!!!! It says find slopes of the functions at x = 1

4. Originally Posted by marie7
omg oops I forgot!!!! It says find slopes of the functions at x = 1
OK. No problem. The slope at any value on a graph is given by the derivative of the function evaluated at that point. So, first we need the derivative. Do you know how to use the chain rule?

ohhhh and then there's the formula y = mx + c

am I onto something here?

6. Originally Posted by marie7
OK. Good. You've found the derivative $y'=18x+6$. (Don't forget the apostophe, marie. It let's peolpe know that this is the derivative.)
Now, all ya gotta do to find the slope is to plug in 1. IE $y'=18(1)+6$. This gives you the slope.

As far as, $y=mx+b$ is concerned, we don't need it here because all we want is $m$. The rest of the equation doesn't concern us. Got it?

7. Cool, gotcha. How about b)?

8. For b) we need the quotient and chain rules.

$
\frac{d}{dx}\left[\frac{2x-5}{(3x+2)^2}\right]=\frac{(3x+2)^2\cdot(2)-(2x-5)\cdot(2)(3x+2)(3)}{[(3x+2)^2]^2}
$

After you simplify, plug in 1.