Sketching Graphs.. I fail. (1 Viewer)

[Doctor Jolly]

I don't understand how to sketch this graph. I read over Maths In Focus's teaching, but I still don't get it. Help?

y = (x + 2) / (x^2 - 4)

 

[morganforrest]

simplify... x^2 - 4 is the difference of two squares. Cross x+2 off the top and bottom then graph y=1/x-2 which you really should know

 

[Doctor Jolly]

simplify... x^2 - 4 is the difference of two squares. Cross x+2 off the top and bottom then graph y=1/x-2 which you really should know

unfortunatley, i don't know that either...
I need help from my teacher, but he doesn't remember how to do these!

Okay, this is what i think you do:
1. Find y intercept
2. Find the limit
3. See if it is a continous or discontinous graph.

Is that right?

 

[morganforrest]

It's an hyperbola that's discontinuous at x=2

 

[morganforrest]

you know its a hyperbola because the top is the derivative of the bottom

 

[Doctor Jolly]

you know its a hyperbola because the top is the derivative of the bottom

yeah, i get that bit...

So the steps that i did before were correct then?

 

[tommykins]

I don't understand how to sketch this graph. I read over Maths In Focus's teaching, but I still don't get it. Help?

y = (x + 2) / (x^2 - 4)

1. Find x/y intercepts (by letting x = 0, y = 0)
2. Find if its an odd or even function ( f[x] = f[-x] -> even or -f[-x] = f[x] -> odd) although this step isn't required.
3. Find assymtotes. The denominator (if it is able to be factorised) can not = 0 as it will make the function undefined. x²-4 factorises to (x-2)(x+2), equating this to 0,your vertical assymtotes are x = 2 and x = -2.
4. Limits for horizontal assymtotes. You divide every term by the HIGHEST powered term in the denominator, in this case- it is x². Whatever is still divided by x becomes 0.

(x/x² + 2/x²) / (x²/x² - 4/x²) = 1/x / 1 = 0 /1 . Thus the horiztontal assytote is y = 0.

5. draw up graph, put in the assymtotes + plot the intercepts (if any).

Now, the shape of the graph can be determined quite easily. You simply test extremities of either side of the VERTICAL assymtotes. since in our case, it is x = 2 and x = -2.

For x = 2, you test x = 1.99999999 into the original equation and start from the given value. Then you test 2.000001 into the originla equation and start from there.

You then test large extremeties (such as -100 and +100) and you'll know the shape of the graph.


EDIT - for your specific graph, you can factorise and cancel out to give 1/(x-2) which is like 1/x but the veritcal assymtote is x = 2.

 

[kurt.physics]

I don't understand how to sketch this graph. I read over Maths In Focus's teaching, but I still don't get it. Help?

y = (x + 2) / (x^2 - 4)

So what do we know? Its a fraction, and as a result, the denominator can not equil zero. So we find out what happens when it does equil zero

x² - 4

The difference of two square is a binomial product (in MIF Prelim Ext 1 text its on page 46)

a² - b² = (a + b)(a - b)

So x² - 4 = (x)² - (2)²
= (x - 2)(x + 2)

Now lets find out when this is zero

(x - 2)(x + 2) = 0

From the null factor, MIF Prelim Ext 1 Page 84

x = 2 , -2

So x can not = 2 or -2

Lets see what happens what happens when x approaches -2

we say, as the limit as x approaches -2

lim(x-->-2) of (x + 2) / (x^2 - 4)

We first try direct substitution, We get

0/0


Now, when its 0 devided by 0, we know that the graph is discontinous at that point. When its a non zero number like 2, 8, -5 devided by 0, there is an asymtope.

So there is a hole at -2

Now if we come back to the function, we can reduce it


(x + 2) / (x - 2)(x + 2)

It becomes

1 / (x - 2)

Now looking at this, there is an asymtope at x = 2 because when we substitute x = 2 into the function we get a non zero number devided by zero.

We can plug in a value into this thats near 2 like 1.999999 to find that

as x-->2 from the left (because is 1.99999), f(x) -->-∞

and as x-->2 from the right (using a number like 2.00000001), f(x) --> ∞



Now if we relook at 1 / (x - 2), we can tell its a hyperbola, so we know what its ruff shape is going to be.

We know

What happens at x = -2

We know what happens at x = 2 from both sides

And we know its going to be a hyperbola

So from our useful detective skills, we have found useful properties of this graph and now can do a sketch of it.

If you want an even more presise picture picture of what the graph would look like, find out what happens when x goes to possitive and negative infinity, you will find it approaches zero.

This is what graphing and sketching is, you find out little important features and details to see where it turns and where it can not go and seeing what it cannot be.

 

[Hankinsd7]

the 1st thing i do is find the limits (because it has an x on the denominator it has to be discontinuous):

x=2 and y=0
because if x=2 then the dinominator is 0 which cannot exist, and because there is no x on the numerator and no constant the fraction cannot equal 0, therefore y=0 is a limit.

my teacher always says the next thing should be to derive and find any turning points. in this case there are none.

next domain and range:

domain: all real x, except x=2 range: all real y, except y=0

find out if its even or odd function, f(x)=f(-x) and f(-x)=-f(x):

i dont really pay attention to this part, but again thats what my teachers say to do.

find intercepts:

y=-1/2

and if all else fails work out a valur table for certain points:

when x=-1 y=-1/3 x=0 y=-1/2 x=1 y=-1 x=3/2 y=-2 x=5/2 y=2 x=3 y=1 x=4 y=1/2


also if u dont know what a certain graph looks like then try using a program to draw them, such as winplot or there are online graphics calculators.

 

[Doctor Jolly]

Now, the shape of the graph can be determined quite easily. You simply test extremities of either side of the VERTICAL assymtotes. since in our case, it is x = 2 and x = -2.

For x = 2, you test x = 1.99999999 into the original equation and start from the given value. Then you test 2.000001 into the originla equation and start from there.

You then test large extremeties (such as -100 and +100) and you'll know the shape of the graph.

Okay, i got everything up until the bolded bit. So do I test x = 2 because it's an assymtote? So does that mean that I also test x = -2?

 

[kurt.physics]

Okay, i got everything up until the bolded bit. So do I test x = 2 because it's an assymtote? So does that mean that I also test x = -2?

Okay, what he means is, as i have said in my post (kurt.physics, post no. 8), we have identified that x = 2 and x=-2 are 'asymtopes', that is, when we are at 2 and -2, freaky shit happens.

Now what tommykins means is that, because something happens, what we should do is see what happens on both sides of the curve. 1.9999999, on the numberline is so so close to 2 from the left but obviously, it isnt. So when we put 1.99999 into the function to find the y value, we get a really big negative number, so the co-ordinates would be (1.99999999, -1000000000000), in other words, as we get closer to 2 from the left direction (side), the graph will curve down heaps.

Now we find out what happens when the x value gets closer to 2 from the RHS (Right Hand Side), one number that is close to 2 but is not and is on the right side of 2 is 2.000000001, we plug it into the equation and see that the curve gets really really high. So we say it goes to infinity.

What tommykins means when we have to plug in large numbers like 100 and -100 is that we are trying to find out what happens when x goes to plus and minus infinity, that is, what happens to the curve when we go to the left and the right of x.

There are two ways of doing so, we can firstly plug in large values into the equation to find out what happens. Lets try x=1,000,000 (a million), if we plug it into the function we get

y = 0.0000000000001

or something like that

So we now know that as x-->∞, the graph gets closer and closer to 0 from the top of the axis because y is positive.

Lets find out what happens when x--> -∞, lets plug in say x = -1,000,000 (negative 1 million)

y = - 0.0000000000001

So when x--> - ∞, the curve approaches the y axis, that is 0, from the bottom side because its negative.

The other way is a bit more complex, it is by using limits

lim(x-->∞) of (x + 2) / (x^2 - 4)

we divide by the x with the highest exponent (the squared one)

So,

x/x² + 2/x² / x²/x² - 4/x²

This becomes

1/x + 2/x² / 1 - 4/x²

Now this way works on the fact that the limit as x-->∞ of 1/x then we can take the limit for each individual 'group' in the equation.

So it becomes

0 + 0/ 1 - 0

This becomes

0/1

which is 0,

Now on the graph of 1/x, as x-->∞ from the top or 'positive' side, then it is 0 from the positive side of the y axis.

You can do the same as for x--> - ∞ but there is one difference, when x-->-∞ on 1/x, it goes to zero from the negative side, so its 0 from the negative side of the axis.

DO NOT WORRY if you dont understand this method, it is quite difficult, it took me a while to fully understand it and im still learning! If your interested, its in the MIF Prelim Ext 1 text page 202-203

Just refer back to my previous post for when x-->-2, its the long one.

Good Luck
Kurt

 

[Doctor Jolly]

Thanks Kurt. I get it now - well most of it anyways.
Thanks again for helping