# Function help

1. Jun 13, 2010

### xX-Cyanide-Xx

This is another question that is impossible for me to complete.i dont even know whereto start.
PLLLease help me.

Draw a sketch of a function with the following properties:

a) The gradient is negative where -2 < x < 2
b) The gradient is positive where x < -2 and where x > 2
c) The gradient of the function is zero at (-2, 1) and (2, -1)
d) The zeros of the function are (-4, 0); (0, 0); and (4, 0)

2. Jun 13, 2010

### Mentallic

Why is this impossible to complete? How can you not possibly know where to start? If you know which way the negative and positive gradients slope, and what zero gradient looks like, and what a zero of the function is, this question is really really easy....
If you don't know what all those things are, you need to revise.

3. Jun 13, 2010

### xX-Cyanide-Xx

Yea, i am really bad at this. i hav been trying this question for hours. can u show me an example or somthing?

4. Jun 13, 2010

### Mentallic

Seriously? Not bragging here but it took me just as long to draw the graph as it did to draw the axis and such.
What are you really bad at? Look at dot points 3 and 4 and they should give you the kick start you need. Draw dots on your graph or something to indicate that those points lie on your function. Since it says at $x=\pm2$ the gradient is zero, draw a short horizontal line through those points to indicate this. Draw a line anywhere between x=-2 and x=2 that is slanting with a negative gradient to indicate which way the graph is going.

But even if you just labelled all the points that you know, it's pretty much as easy as connect-the-dots.

5. Jun 13, 2010

### xX-Cyanide-Xx

ive had a closer . look, but im still a little confused about hoe to draw the graph. how do i know wat shape is it? like is it cubic, or Quadratic.. so on.

I'm really sorry for being annoying but im a very slow learner. but i appreciate your help.

6. Jun 13, 2010

### Mentallic

Look at the last dot point again.

7. Jun 13, 2010

### xX-Cyanide-Xx

i had a look at them, i even tried graphing them, but it just makes a straight line. you said that u drew a graph before. could you please show me that graph and tell me which points are which?

8. Jun 13, 2010

### Mentallic

You tried using a graphics calculator to see what it comes up with when you enter in the 3 zeroes (-4, 0), (0, 0), and (4, 0) ???

What do you mean by "tell me which points are which"? Why don't you put that graphics calculator down for a second, include the info from dot point 3 on your graph (which you should have drawn up on paper by now) and then, even though the answer should be clear from this point, look at dot points 1 and 2 for clarification.

9. Jun 13, 2010

### xX-Cyanide-Xx

i had another look. should the graph look like this? the picture is in the attachment. i made it in paint.

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• ###### EmptyGraph.gif
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10. Jun 13, 2010

### xX-Cyanide-Xx

that graph is just the points from part 3 and 4 of the question. then i connected the dots.

is this correct???

11. Jun 13, 2010

### Mentallic

I can't see attachments until they're approved. Does the graph look like a quadratic, or a cubic, or a quartic etc.?

12. Jun 13, 2010

### Mentallic

As long as it's a smooth curve, with no sharp turns or anything, then yes it's correct.

13. Jun 13, 2010

### xX-Cyanide-Xx

yay,thats wat i did, and it came out to be a cubic graph. so thats correct?

can u check later once the attachment is approved just to check and make sure?

ur even better than my maths teacher.

14. Jun 13, 2010

### Mentallic

Yes it's a cubic. This should be very clear by the fact that there are 3 zeroes to this function.
Now, I didn't want to bring this up earlier because it would've made things more complicated without them needing to be, but you might be interested to know that there are no real cubics that can have have the info you provided. It's not possible to make a cubic have zeroes -4,0,4 AND turning points at $x=\pm 2$.

Also, even though it says that for x>4 and x<-4, the gradient is positive, this doesn't necessarily mean for large positive x, the function becomes very large and for large negative x, the function becomes very largely negative. You can have things like horizontal asymptotes which the function approaches but never reaches, and still have positive gradient.

15. Jun 13, 2010

### Mentallic

Ok, looking at your attachment, I would agree that it's not possible to determine exactly what happens at x>4 and x<-4, but I'd assume that the function is continuous for all x, so you should extend the graph so that it keeps going on like all cubics do. All you have to do is extend the lines at either end on the positive and negative side and put arrows on the end of them to show that it keeps going.

16. Jun 13, 2010

### xX-Cyanide-Xx

but wat do i do for these parts:
a) The gradient is negative where -2 < x < 2
b) The gradient is positive where x < -2 and where x > 2

17. Jun 13, 2010

### Mentallic

Umm... Do you know which way a line with positive gradient slopes, and which way a negative gradient line slopes? Well all you do is make sure your curve is sloping the correct way in those regions.

18. Jun 13, 2010

### xX-Cyanide-Xx

I kinda get where ur at, it would be easier if i new wat the question ment in more detail. so wat does this symbol mean in this question: <. and wat does the question actually ask.(explain every detail anout wat this,-2 < x < 2. and this means, x < -2 and where x > 2.)

19. Jun 13, 2010

### Mentallic

You don't know what < and > mean? This is basic stuff learnt in primary school. x<-2 reads "x is less than -2" and -2<x<2 reads "-2 is less than x is less than 2" so this is saying all values of x between -2 and 2.

20. Jun 13, 2010

### Redbelly98

Staff Emeritus
(a) Check the slope (gradient) of your graph between -2 < x < 2. Does the slope appear to be positive or negative there?