Curve sketching (1 Viewer)

[mtsmahia]

Hi guys,

I had a question on the curve y=x^2/x^2 -9, or any curve in that form

when i graph it, it come out to be 2 hyperbola segments in the first 2 quadrants and a concave down parabola in the bottom...

So my Q is, do all curves like this have an concave down shape?

and how do you recognize the changes in shape/transformations of this overall graph.

And btw im in Year 11, and we havent learnt critical points and stuff, we have just been graphing using asymptotes, domain,range and intercepts.
sry if u found it hard to understand!
thanks

 

[Timothy.Siu]

Hi guys,

I had a question on the curve y=x^2/x^2 -9, or any curve in that form

when i graph it, it come out to be 2 hyperbola segments in the first 2 quadrants and a concave down parabola in the bottom...

So my Q is, do all curves like this have an concave down shape?

and how do you recognize the changes in shape/transformations of this overall graph.

And btw im in Year 11, and we havent learnt critical points and stuff, we have just been graphing using asymptotes, domain,range and intercepts.
sry if u found it hard to understand!
thanks

well like you said, just use asymptotes, and intercepts
u shudn't really do these questions by memory although i guess u can have a general idea of what the answer will be.

 

[mtsmahia]

But how do u know if what type of graph to draw in the quadrants, i mean; how do u deduce that from the Function

 

[kurt.physics]

Hi guys,

I had a question on the curve y=x^2/x^2 -9, or any curve in that form

when i graph it, it come out to be 2 hyperbola segments in the first 2 quadrants and a concave down parabola in the bottom...

So my Q is, do all curves like this have an concave down shape?

and how do you recognize the changes in shape/transformations of this overall graph.

And btw im in Year 11, and we havent learnt critical points and stuff, we have just been graphing using asymptotes, domain,range and intercepts.
sry if u found it hard to understand!
thanks

The whole point of curve sketching is to determine special characteristics of functions and graph them with this information. So instead of memorizing all different types of graphs, you should acquaint yourself with the knowledge and understanding of how to figure out what a function looks like in the case you get a weird one.

So when graphing these kind of functions the first thing you should look at is asymptotes first. Find where they are, if they are vertical or horizontal and what the curve approaches when they get near the asymptotes.

Are you doing 2 unit mathematics or 3 unit mathematics. There is a 3 unit mathematics technique for determining if the graph is above or below the x-axis.

When you are in year 12 is where you have to start worrying about stationary points, maximum, minimum, concavity, inflections etc. All these things will add to your graphs, but for year 11 it is only really neccesary to know the things you mentioned ie asymptotes, intercepts, domain and range etc.

 

[mtsmahia]

The whole point of curve sketching is to determine special characteristics of functions and graph them with this information. So instead of memorizing all different types of graphs, you should acquaint yourself with the knowledge and understanding of how to figure out what a function looks like in the case you get a weird one.

So when graphing these kind of functions the first thing you should look at is asymptotes first. Find where they are, if they are vertical or horizontal and what the curve approaches when they get near the asymptotes.

Are you doing 2 unit mathematics or 3 unit mathematics. There is a 3 unit mathematics technique for determining if the graph is above or below the x-axis.

When you are in year 12 is where you have to start worrying about stationary points, maximum, minimum, concavity, inflections etc. All these things will add to your graphs, but for year 11 it is only really neccesary to know the things you mentioned ie asymptotes, intercepts, domain and range etc.

Got it, but as i said before, how do you know the actual shape of the function by analysing the function. I can find the asymptotes, domain/range but when it comes to actually sketching the curve i get confused.. specially with this one....how do you know it is a concave down parabola? & and why are there two hyperbolic curves in the first two quadrants

oh yeah...and im doing 3 unit

 

[Aquawhite]

That would be one freaky looking graph :S...

 

[kurt.physics]

Got it, but as i said before, how do you know the actual shape of the function by analysing the function. I can find the asymptotes, domain/range but when it comes to actually sketching the curve i get confused.. specially with this one....how do you know it is a concave down parabola? & and why are there two hyperbolic curves in the first two quadrants

oh yeah...and im doing 3 unit

So it easy to see the vertical asymptotes occur when x² - 9 = 0 ie when x = 3 and x = -3

So draw the dotted lines x = 3 and x = -3

Using your calculator it is easy to see that the curve approaches + infinity to the left of -3 and negative infinity to the right of -3. And also the curve approaches negative infinity to the left of 3 and approaches + infinity to the right of 3.

Also check for the x and y intercepts.

For x intercepts, y = 0. ie



When x = 0, y = 0. Therefore (0,0) is the intercepts.

This means that the curve does not cross the axis at any other point.

Also, you may check that y = 1 is the horizontal asymptote.

So we know that the curve aproaches 1 from above when y becomes very large, and as we get closer to x=3, the curve aproaches infinity. So we draw a curve similar to that of a hyperbola (if you get what i mean =)

See if you can use the same logic for the rest and see if that works

 

[Iruka]

I think the horizontal asymptote is y=1.

 

[omniscience]

1+1 =2

 

[Iruka]

Prove it.

 

[omniscience]

Prove it.

no u

 

[kurt.physics]

I think the horizontal asymptote is y=1.

ah, yes, sorry ... Illogical false presumption on my behalf =S

 

[Uncle]

Graph of [maths]\frac{x^{2}}{x^{2}-9}[/maths]

When plotting functions in fractional form, the value of the asymptote makes the denominator equal zero.

 

[mtsmahia]

Graph of [maths]\frac{x^{2}}{x^{2}-9}[/maths]

When plotting functions in fractional form, the value of the asymptote makes the denominator equal zero.

isnt there meant to be a concave down parabola at the bottom? cuz thats what ym textbook shows

 

[Aquawhite]

isnt there meant to be a concave down parabola at the bottom? cuz thats what ym textbook shows

Why would there be? It looks like a hyperbole questions to me? unless I'm mistaken (then again, I'm only briefly looking at this question)

 

[khorne]

Kurt's method is correct, except it only works for rational functions (poly/poly) like that...It works under the premise that the only place a curve can change signs is at an intercept or discontinuity.

Also, a hint if anyone needs it:

The horizontal asymptotes only occur when the denominator is 0 while the numerator is a number (not 0). Eg x-3/((x-3)(x+3)) only has an asymptote at x = -3, while x = 3 is a discontinuity (as 0/0 is undefined)

 

[cutemouse]

The horizontal asymptotes only occur when the denominator is 0 while the numerator is a number (not 0).

ie. undefined.

(as 0/0 is undefined)

No, that is mathematically indeterminable.

 

[khorne]

Ok..terminology is not correctly applied...You bump a month old threadfor that?

 

[blaze.bluetane]

Note that any real function composed of a fraction can be converted into a rational/rational function form. Otherwise is should not have any points of discontinuity an terms of domain

Looking at the equation, upon factorisation of the denominator, it can be seen that for the function to exist, x must not equal 3 or -3. Therefore horizontal asymptote at both of these points

As x approaches infinity, y approaches 0 from the positive side
Similarly, as x approaches negative infinity, y approaches 0 from the positive side

As x approaches 3^-, y approaches negative infinity
As x approaches 3⁺, y approaches positive infinity

As x approaches -3^-, y approaches positive infinity
As x approaches -3⁺, y approaches negative infinity

Obviously, intercepts at (0,0)

If you test some point a little less than -3 and a little more than 3 you can see that the curve is smooth
Similarly, between -3 and 3, there is a smooth, CONICAL curve

So plot these points (sections) on your graph and draw smooth curves between them. The concave down conical curve is the smooth curve between limits at plus +-3 and (0,0)

So sketching (notice that Uncle's curve is a bit 'zoomed out'), there is a mixture between a hyperbola and a parabola between 3 and -3 (apex and limits approaching infinity) and there is two hyperbolic arms above the x axis.


Oh and the hyperbolic arms are just found by finding the asymptotes and limits and sketching smoothly

 

[blaze.bluetane]