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No idea where to start please explain thank you very much

Dec 17, 2018

#1
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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
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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
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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
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ahhh ok thanks for the details ill give it a go now

YEEEEEET  Dec 17, 2018
#5
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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
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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)

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
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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
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yes LOL

Melody  Dec 17, 2018
#23
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That's some keen insight there Melody!

Rom  Dec 17, 2018
edited by Rom  Dec 17, 2018
edited by Rom  Dec 17, 2018
#24
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Thanks Rom

Melody  Dec 17, 2018
#10
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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
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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
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I just took a photo of some of my scribble, it is not to scale.

Dec 17, 2018
#13
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ah so basically the x axis of the two points

thank you for using ur time to teach me!!

YEEEEEET  Dec 17, 2018
#14
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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
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I agree with Melody.....these ARE confusing....!!!!

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

CPhill  Dec 17, 2018
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I shall grind these type of question tomorrow ;) thanks for the advice

YEEEEEET  Dec 17, 2018
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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
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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...

CPhill  Dec 17, 2018
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No one likes my #6.              I was offering a real insight

Dec 17, 2018
#20
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Mmmmmm...  I liked it.....

CPhill  Dec 17, 2018
#21
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and liked it shall be

YEEEEEET  Dec 17, 2018
#22
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You do not sound convinced LOL :)

Melody  Dec 17, 2018