#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