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avatar+838 

No idea where to start please explain thank you very much

 Dec 17, 2018
 #1
avatar+701 
-1

First, understand what an asymptote is. An asymptote is "a line that continually approaches a given curve but does not meet it at any finite distance." - google 

 Dec 17, 2018
 #2
avatar+838 
+1

yes i have already learned about asymptotes, what i struggle with is the actual question of sketching the gradient function...

YEEEEEET  Dec 17, 2018
 #3
avatar+100813 
+2

These can be quite tricky.

 

consider the first one.

when x=1/6 the gradient is 0

when x=-3/2 the gradient is 0.

plot those 2 points first.

 

now when x>1/6 the gradient is negative and it is getting steeper and steeper.

so as x tends to infinity, the gradient tends to -infinity.  

 

When x is <-2/3 the same thing is happening

so as x tends to -infinity, the gradient tends to -infinity also.

 

Now between -2/3 and 1/6 the curve is increasing.  I mean the gradient of the tangent is positive

Somewhere in the middle between the 2 is a point of inflection.  This is where the first derivative is still positive but it is becoming smaller so the graphe between -2/3 and 1/6 will look a bit like a concave down parabola. the point at the top will be the point of inflection of the original function graph.

 

Do you get all that?  Try drawing it.  I think these are fun to work out.

 

Then have a go at the second one.

 

Partial mathematician:

I know you are keen to learn and I really do appreciate that, I want you to stick around for a long time ....  but please stop pretending that you know a whole lot more than you really do.  You leave people confused and sometimes annoyed. Some of your input is really good, I know this, but ....  you can finish this sentence for yourself.

 Dec 17, 2018
edited by Melody  Dec 17, 2018
 #4
avatar+838 
+1

ahhh ok thanks for the details ill give it a go now 

YEEEEEET  Dec 17, 2018
 #5
avatar+838 
0

btw did you mean -3/2 instead of -2/3?

 

Yes I did, sorry.  (edit by Melody )

YEEEEEET  Dec 17, 2018
edited by Melody  Dec 17, 2018
 #6
avatar+100813 
+2

If you think about it this first one is a cubic graph (degree of 3)     (up down and up again)

When you differentiate a cubic you get a quadratic (with a degree of 2)

Quadratics are parabolas   :)

 

So the graph of the gradients of a cubic will give a parabolic curve.

 

\(y=x^3 \qquad \text{cubic}\\ y'=3x^2 \qquad \text{parabolic}\\ y''=6x \qquad \text{Straight line}\\ y'''=6 \qquad \text{horizonal line}\\ y''''=0 \\\)

 

see if you can work out the relevance of all that. :)

Melody  Dec 17, 2018
 #8
avatar+838 
0

so for the first one so far i tried to make -3/2 be one of the first x intersect and im assuming it crosses the y intersect at 9?? and the shape so far is a negative quadratic.

 

The site is playing up and will not let me respond as usual so I will try and respond here.

 

We cannot know where it will cross the y axis. We are not given enough info for that. there are no numbers even on the original y axis, not that that would help much.  The -9 has got nothing to do with this. We are only interested in the x values and the slope.

YEEEEEET  Dec 17, 2018
edited by YEEEEEET  Dec 17, 2018
edited by Melody  Dec 17, 2018
 #8
avatar+100813 
0

yes LOL

Melody  Dec 17, 2018
 #23
avatar+5076 
0

That's some keen insight there Melody! cool

Rom  Dec 17, 2018
edited by Rom  Dec 17, 2018
edited by Rom  Dec 17, 2018
 #24
avatar+100813 
0

Thanks Rom   wink

Melody  Dec 17, 2018
 #10
avatar+100813 
+1

I added the answer on you question on your post because the site was plying up. Hopefully it will behave itself now.

 Dec 17, 2018
 #11
avatar+838 
0

A wildguess, so by using the two coordinates i can calculate the gradient and by that i should be able to work out the x intersection between the two points which mean i can workout my other intersection in my new graph?

YEEEEEET  Dec 17, 2018
 #12
avatar+100813 
+2

I just took a photo of some of my scribble, it is not to scale.

 

Hopefully it will help you.

 

 Dec 17, 2018
 #13
avatar+838 
+1

ah so basically the x axis of the two points

thank you for using ur time to teach me!!

YEEEEEET  Dec 17, 2018
 #14
avatar+100813 
+2

I hope you understand but you will need to practice to cement it into your brain.

 

Also you need to think about some of the sideways things I have talked about.  (Mainly in #6 post)

If you can work out their relevance it will help you a real lot.

Melody  Dec 17, 2018
edited by Melody  Dec 17, 2018
 #15
avatar+100519 
+2

I agree with Melody.....these ARE confusing....!!!!

 

Her last graph is pretty good....it will be impossible to create an "exact" graph

 

cool cool cool

CPhill  Dec 17, 2018
 #16
avatar+838 
+2

I shall grind these type of question tomorrow ;) thanks for the advice

YEEEEEET  Dec 17, 2018
 #17
avatar+100813 
+2

Thanks Chris,

If you  draw them one under the other so that the x values line up, it help with the graphing too.

I sort of did that. 

 Dec 17, 2018
 #18
avatar+100519 
+2

Yep....that's the best way to "see" these, I think

 

There is so much Calculus that I have forgotten......good to get a "refresher"  here...

 

 

 

cool cool cool

CPhill  Dec 17, 2018
 #19
avatar+100813 
+1

No one likes my #6.     sad         I was offering a real insight     

 Dec 17, 2018
 #20
avatar+100519 
+2

Mmmmmm...  I liked it.....

 

 

cool cool cool

CPhill  Dec 17, 2018
 #21
avatar+838 
+1

and liked it shall be

YEEEEEET  Dec 17, 2018
 #22
avatar+100813 
0

You do not sound convinced LOL :)

Melody  Dec 17, 2018

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