Cambridge 3U - Chapter 3 Simple Hyperbola question (1 Viewer)

[deswa1]

First off, you want to graph y=(x-4)/(x+2). There are a few ways to do this. Personally, I would realise that there is no value at x=-2 because the denominator becomes zero. At this point approaching from the negative side, y approaches infinity and approaching from the positive side, y approaches negative infinity. Also, as x approaches plus or minus infinity, y approaches one which is a horizontal asymptote. Finally, When y=0, x=4 which is the x intercept and when x=0, y=-2 which is the y intercept. Use this information (plus the general shape of the hyperbola) to graph it. Then simply test values 'inside' and 'outside' the hyperbola to work out what areas satisfy the inequality.

Another way to graph the function is as follows:
y=(x-4)/(x+2)=(x+2-6)/(x+2)
y=1-(6/x+2)

NOTE: If all you wanted to do was solve the inequality, it would be much easier to do it algebriacally (all less than signs should be less than or equal to):
(x-4)/(x+2)>0
(x-4)(x+2)>0 on multiplying by (x+2)^2
x>4, x<-2

 

[D94]

Look for:
1. Vertical Asymptotes
2. Horizontal Asymptotes
3. Oblique Asymptotes
4. x and y intercepts
5. Left hand and right hand limits for the Vertical Asymptotes
6. Check if function crosses an asymptotic line

Then, some questions ask for max/min and inflexions etc.; these can be done with differentiation.

 

[D94]

So essentially, for the function, f, above:

VA: x =/= -2
HA/OA: 1
y = 0; x = 4
x = 0; y = -2

limit as x tends to -2 (+) = negative value
limit as x tends to -2 (-) = positive value

Sketch: looks like the -1/x graph but raised up by 1 and moved to the left by 2.

 

[D94]

Look for:
3. Oblique Asymptotes - um not sure what you mean, you mean the general shapes of the hyperbola?
5. Left hand and right hand limits for the Vertical Asymptotes - only learned a few limits like natural domain and cannot be -sqrt, dont think we've learned anything else
6. Check if function crosses an asymptotic line - could you explain what you mean by this?

Then, some questions ask for max/min and inflexions etc.; these can be done with differentiation. --- haven't learned at all :S

Ok:

You have vertical asymptotes, horizontal asymptotes and you can have oblique/slanted/diagonal asymptotes. This can be found using polynomial long division or factorisation, but I guess you may have not learned that. You probably won't need to know this for MX1 right now.

For limits, you look as x approaches the left hand side of the vertical asymptote, and see if it goes towards positive or negative infinity. And likewise with the right hand side. This is like when you look for what happens when x tends to +/- infinity, but instead, as x approaches a value.

An intuitive definition of an asymptote is where the function approaches that value but never crosses it. However, it can, because an asymptote is only ever really applicable for very large or very small values of x. So, you need to check if the function crosses these "asymptotes".

But, you say you haven't even done differentiation; what year are you in? :/

 

[D94]

Thanks, I know a way (i think it's quite dodgy) but not the proper way to get horizontal asymptote? Can you please tell me the way you find the horizontal asymptote?

Divide through by the highest power of 'x'.

 

[D94]

I'm in year 11 doing mx1. Btw sorry but when you say "Divide through by the highest power of 'x'. ", highest power of x is 1 in this equation but what do I divide through? sorry, i'm being a bit vague, let me find the exact question and see if it gives any more info.

Oh, Year 11. I see; then I guess we'll revert back to basics.

For this function, firstly, look at the intercepts, then look at the limits and asymptotes. To find the horizontal asymptote, you divide through by the highest power of x, so this in this case, it becomes: (x/x - 4/x)/(x/x + 2/x) = (1 - 0)/(1 + 0), because as x tends to infinity, 4/x and 2/x tends to zero.

I think you should just make sure your textbook has covered this; I don't recall any other method of doing so at a Year 11 stage :s

 

[D94]

The exact topic we're studying is "Intercepts and Signs) [in Graphs and Inequation] and this the what the question is asking:

If necessary, collect all terms on the LHS and factor. Then solve the inequation by finding any zeroes [x-intercepts] and discontinuities and drawing up a table of values around them.

x - 4 / x + 2 is < or = 0

Oh I see, yeah, refer to deswa1's post then.

Another intuitive way to find the horizontal asymptote for this question is to say, if x gets very large, the numerator and denominator both grow almost at a constant rate, so if x = 1000, then it's 996/1002, if x is 10000, then it's 9996/10002, so it's clear that as x gets larger, the functions tends to 1.

 

[deswa1]

I'll give an example, just wait for me to type it up...

<a href="http://www.codecogs.com/eqnedit.php?latex=f(x)=\frac{x-4}{x@plus;2}\\ f(x)= \frac{\frac{x}{x}-\frac{4}{x}}{\frac{x}{x}@plus;\frac{2}{x}}\\\\ \textup{On dividing everything through by x}\\ f(x)=\frac{1-\frac{4}{x}}{1@plus;\frac{2}{x}}\\ \lim_{n \to \infty }\frac{4}{x}=\lim_{n \to \infty }\frac{2}{x}=0\\ \therefore \lim_{n \to \infty }f(x)=\frac{1}{1}=1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?f(x)=\frac{x-4}{x+2}\\ f(x)= \frac{\frac{x}{x}-\frac{4}{x}}{\frac{x}{x}+\frac{2}{x}}\\\\ \textup{On dividing everything through by x}\\ f(x)=\frac{1-\frac{4}{x}}{1+\frac{2}{x}}\\ \lim_{n \to \infty }\frac{4}{x}=\lim_{n \to \infty }\frac{2}{x}=0\\ \therefore \lim_{n \to \infty }f(x)=\frac{1}{1}=1" title="f(x)=\frac{x-4}{x+2}\\ f(x)= \frac{\frac{x}{x}-\frac{4}{x}}{\frac{x}{x}+\frac{2}{x}}\\\\ \textup{On dividing everything through by x}\\ f(x)=\frac{1-\frac{4}{x}}{1+\frac{2}{x}}\\ \lim_{n \to \infty }\frac{4}{x}=\lim_{n \to \infty }\frac{2}{x}=0\\ \therefore \lim_{n \to \infty }f(x)=\frac{1}{1}=1" /></a>