We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

#1**-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

PartialMathematician Dec 17, 2018

#3**+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.

Melody Dec 17, 2018

#6**+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**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

#10**+1 **

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

Melody Dec 17, 2018

#12**+2 **

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

Hopefully it will help you.

Melody Dec 17, 2018

#13**+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**+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