Revision [Question and Answers Provided] Need help understanding ASAP

Bob · 2011-03-09 00:59:08

Getting there.

I've got to go out; I'll leave you in bobbym's very capable hands. Back later

Bob

bobbym · 2011-03-09 00:59:44

Hi;

That is correct we write that more conveniently as:

Now please do this one on your own.

Alive · 2011-03-09 01:08:42

later bob bundy

so i followed the steps and got

2 - (√6/-2)

Right?

I still don't know what this means though...

Last edited by Alive (2011-03-09 01:09:21)

bobbym · 2011-03-09 01:12:42

Hi;

That is correct.

We write like this because minus a minus is a plus:

Now what do you do not understand?

Alive · 2011-03-09 01:16:22

Where to plot this in the x axis when trying to sketch

bobbym · 2011-03-09 01:23:14

Hi;

These 2:

represent where the curve crosses the x axis. We found them by setting y to 0.

To find where it crosses the y axis you set x to 0 in the function which I think is

We already have the answers to that in an earlier post.

Alive · 2011-03-09 01:28:56

Yeah but... as in....
What does 2 - √6 /2 actually mean? I know it's strange but... I don't know where to plot it even though I've got that answer... It reads 2 take away √6 which is divided by 2? How will we know where that is in the graph?

Bob · 2011-03-09 02:48:55

hi Alive

For a sketch you don't have to do an accurate plot. Just show roughly where the point is.

Now root 6 must be between 2 and 3 so half it must be beween 1 and 1.5.

So when you take from 2 you get a small positive value and when you add it to 2 you get a value just over 3.

That's enough for a sketch. Draw axes, mark one crossing point at just over zero and another at just over 3.

You've already got y = -5/4 as the crossing point on the y axis so put a point on for that.

Now, what about asymptotes?

Any line that the curve approaches but doesn't get to.

Think about what happens when x = 2.

Also what happens when x tends to infinity and also when x tends to minus infinity.

These three things leads to three asymptotes. (but actually two are the same line)

Can you find them all ?

(In an earlier post I referred to a set of steps to go through for a sketch. Have a look at that for the general principles and some exercises.)

Bob

Last edited by Bob (2011-03-09 02:51:12)

Alive · 2011-03-09 03:46:35

Thanks for the help bob bundy, makes sense now and yeah I know how to get the asymptotes.

I need to know the steps to these before the test.. If when available and whichever you can, please do these (not much time left so I just need to quickly know the steps to questions like these textbook questions..

Sketch each of the following, clearly showing the position of the asymptotes and the intercepts with the axes (correct to 1 decimal place where appropriate).

Sketch the graphs of each of the following.

Thank you

Last edited by Alive (2011-03-09 03:47:24)

Bob · 2011-03-09 07:36:44

hi Alive

(i) Where does the curve cross the x axis?

y = 0 so

I'll leave you to calculate that.

(ii) Where does the curve cross the y axis?

Put x = 0

Arhh! Cannot do that as y zooms off to infinity. Note also, it doesn't matter if we let x approach zero from small positive or small negative values. The squaring means the term is big and positive and taking 3 away makes no difference. y tends to + infinity.

The line x = 0 (y axis) is an asymptote.

(iii) How does y behave when x tends to infinity? How does y behave when x tends to - infinity.

As x tends to + or - infinity, the fraction tends to zero (from positive values) so y tends to -3 from above.

The line y = -3 is another asymptote.

(iv) Does the curve have turning points? Are they maxima, minima or points of inflexion? What is the y value for each x; or, at least, is y positive or negative in these cases.

There are no values of x that will make this zero (but note that the gradient will tend to zero as x tends to + or - infinity.

(v) What symmetry does the curve have? (in an axis for example or about the origin by rotation) Do you know about odd and even functions?

All powers of x are even so the curve has the y axis as a line of symmetry. If you haven't met this idea before here's why I can say that.

Choose any values for x (say x = 10) . y = -2.995. Now work out y when x = -10. It is y = -2.995. That happens for every values of x. So if (a, b) lies on the curve so does (-a, b) => the y axis is a line of symmetry.

(vi) Does the curve have any asymptotes?

Well I have found two already. Are there any more? Answer no. Because I've done the "tends to infinity" bit and the only x value that makes y tend to infinity is zero. So I've got all the elements I need to make a sketch.

I've put this below.

Bob

Last edited by Bob (2011-03-09 07:42:03)

Bob · 2011-03-09 10:04:05

hi Alive

Ok. A quick trip through the others.

x = 0 gives a + fraction for y.
y=0 leads to no solutions to the quadratic as you need sq rt of a negative.
x = 2 is an asymptote as y tends to + infinity as x tends to 2 from either side.
x tends to + or infinity leads to second fraction tends to zero so y tends to 2/3 from above.

next one

I'll do y = 1/x first then Ill modify it.

x = 0 and y = 0 are not possible. Both axes are asymptotes.

This is an odd function. This means if (a,b) lies on the curve then (-a,-b) does also.
So the curve is symmetrical about the origin => first quadrant section may be rotated 180 to get the third quadrant section.

The absolute value will flip the third quadrant section into the second quadrant.
Subtracting ¾ will move the whole graph down by that amount.

last one

Consider 1/xsquared 1 first then apply the absolute value.

y = 0 => x = + or -1.

x=0 is not possible but is an asymptote. y tends to infinity as x approaches zero from either side.

As x tends to + or infinity y tends to -1 so this is an asymptote.

Sketch this then flip the negative bits.

sketches below

Bob

Last edited by Bob (2011-03-09 10:23:54)

Alive · 2011-03-10 00:43:40

Thanks guys, I went well in the test

bobbym · 2011-03-10 00:55:14

How did you do?

Alive · 2011-03-10 00:58:50

I thought questions were going to be harder but it was just kind of the general things like find the point of intersection, turning point.. binomial theorem.. transformation of graphs.. + absolute value... It was pretty easy. Next test will be harder I guess with modules, matrices etc..

bobbym · 2011-03-10 00:59:59

Congratulations! I am glad to hear it was for you!

Bob · 2011-03-10 01:02:14

hi Alive,

That's great. Thanks for letting us know.

Bob

Alive · 2011-03-10 01:02:16

Thanks bobbym and bob bundy, with the help of your answers it was really easy to overcome the difficulties

Last edited by Alive (2011-03-10 01:02:46)

bobbym · 2011-03-10 01:03:53

Bob helped a lot.

Alive · 2011-03-10 01:09:21

^+1

Math Is Fun Forum

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#36 2011-03-09 10:04:05

Re: Revision [Question and Answers Provided] Need help understanding ASAP

#37 2011-03-10 00:43:40

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#38 2011-03-10 00:55:14

Re: Revision [Question and Answers Provided] Need help understanding ASAP

#39 2011-03-10 00:58:50

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#40 2011-03-10 00:59:59

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