# Help on Easy Slope Problems

• May 13th 2009, 07:58 PM
lastlion
Help on Easy Slope Problems
Hello, this is my first post. I am trying to learn some easy algebra and was hoping to get some help.

I have some problems, and answers, but no idea how to arrive at the answers.

1) Find the slope of the following functions at X = 3

Q: Y = 4 + 3X^2
A: Slope = 18

Q: Y = 5X = 6X^3
A: 167

Q: Y = 6X
A: 6

For some reason the x^2 (x squared) is throwing me off. This is homework and answers from an old class, long time ago. Can't remember how to work it.

THANKS
• May 13th 2009, 08:17 PM
Chris L T521
Quote:

Originally Posted by lastlion
Hello, this is my first post. I am trying to learn some easy algebra and was hoping to get some help.

I have some problems, and answers, but no idea how to arrive at the answers.

1) Find the slope of the following functions at X = 3

Q: Y = 4 + 3X^2
A: Slope = 18

Q: Y = 3X = 6X^3
A: 167

Q: Y = 6X
A: 6

For some reason the x^2 (x squared) is throwing me off. This is homework and answers from an old class, long time ago. Can't remember how to work it.

THANKS

If you've taken calculus, the easiest way to do this is by differentiating each function and then evaluate it at $\displaystyle x=3$.

For example, in the first problem $\displaystyle y=4+3x^2$. It follows that the derivative is $\displaystyle \frac{\,dy}{\,dx}=6x$, which gives us the general equation for the slope. Now to find the slope at 3, we plug it into our derivative to get $\displaystyle \left.\frac{\,dy}{\,dx}\right|_{x=3}=6\cdot3=18$.

Try to do something similar for the other two. Just keep in mind that if $\displaystyle y=x^n$, then $\displaystyle \frac{\,dy}{\,dx}=nx^{n-1}$.

Does this make sense? If not, post back and I'll see if we can do it a different way. :D
• May 13th 2009, 08:26 PM
I-Think
Calculus
Buenas noches. Well lastlion, these are not algebra questions, but rather calculus. Basic differentiation skills are required to work these questions.
When you differentiate an equation/function, you can find its slope at any particular point

Q1:
$\displaystyle Y = 4 + 3X^2$
$\displaystyle \frac{dY}{dX}=6X$
$\displaystyle Let X=3$
$\displaystyle Slope=18$

Q3:
$\displaystyle Y = 6X$
$\displaystyle \frac{dY}{dX}=6$
$\displaystyle Slope=6$

Q2:
$\displaystyle Y = 3X = 6X^3$
Needs clarification.
Point 1:If $\displaystyle Y=3X$, then the slope is $\displaystyle 3$, and if $\displaystyle Y=6X^3$, the slope is $\displaystyle 162$. Neither of which is your answer
Point 2: This is in the form of an equation to be solved for $\displaystyle X$,in which case the solution would be $\displaystyle X=\pm\sqrt{\frac{1}{2}}$.
I am very confused...:confused:

First time using a smiley in a post(Smirk)
(2nd time now.)

Edit: It seems I'm a bit too late, Chris L T521 is on the scene
• May 13th 2009, 08:28 PM
lastlion
Chris L.
I was trying to use simple algebra. Rise over run type approach. Maybe that wasn't correct way to think.

I will look in to learning how to find the derivative. I need to learn this the correct way, so that I can apply it to different problems.

Might take me a while. I will try to post back tommorrow to if I have learned the way you suggest or have questions.

Thanks.
• May 13th 2009, 08:32 PM
lastlion
Quote:

Originally Posted by I-Think
Buenas noches. Well lastlion, these are not algebra questions, but rather calculus. Basic differentiation skills are required to work these questions.
When you differentiate an equation/function, you can find its slope at any particular point

Q1:
$\displaystyle Y = 4 + 3X^2$
$\displaystyle \frac{dY}{dX}=6X$
$\displaystyle Let X=3$
$\displaystyle Slope=18$

Q3:
$\displaystyle Y = 6X$
$\displaystyle \frac{dY}{dX}=6$
$\displaystyle Slope=6$

Q2:
$\displaystyle Y = 3X = 6X^3$
Needs clarification.
Point 1:If $\displaystyle Y=3X$, then the slope is $\displaystyle 3$, and if $\displaystyle Y=6X^3$, the slope is $\displaystyle 162$. Neither of which is your answer
Point 2: This is in the form of an equation to be solved for $\displaystyle X$,in which case the solution would be $\displaystyle X=\pm\sqrt{\frac{1}{2}}$.
I am very confused...:confused:

First time using a smiley in a post(Smirk)
(2nd time now.)

Edit: It seems I'm a bit too late, Chris L T521 is on the scene

I think,
I wrote the one problem wrong. That explains your confusion.
Let X = 3
Q: Y = 5X = 6X^3
A: 167

I think I need to be looking at Calculus. I am trying to prepare for a class that starts this summer. I looked at the problem and just assumed algebra. Bad assumption I think.

Your explanations have been very helpful. I think I at least know where to start. Thanks.