Curve Sketching (1 Viewer)

[ronnknee]

I don't think I've learnt to sketch curves properly so can anyone help me out?

Here's my understanding:

Asymptotes: for the horizontal ones, do you just divide everything by x with highest power; for the vertical ones, the denominator cannot be zero so the asymptotes are those numbers which make it zero

Intercepts: let x be zero to find y intercept, let y be zero to find x intercept

If what I said is true,
then for y = 1 / (x^2 - 4)
the horizontal asymptote is y = 0
the vertical asymptotes are x = 2, x = -2
the y intercept is 1
the x intercept is -1/4
the curve would look like a cubic except they never touch x = 2 or x = -2
Is this right?

And what about these:
y = 2x / (x^2 -4)
y = (x^2 + 1) / (x^2 - 4)

 

[watatank]

I don't think I've learnt to sketch curves properly so can anyone help me out?

Here's my understanding:

Asymptotes: for the horizontal ones, do you just divide everything by x with highest power; for the vertical ones, the denominator cannot be zero so the asymptotes are those numbers which make it zero

Intercepts: let x be zero to find y intercept, let y be zero to find x intercept

If what I said is true,
then for y = 1 / (x^2 - 4)
the horizontal asymptote is y = 0
the vertical asymptotes are x = 2, x = -2
the y intercept is 1
the x intercept is -1/4
the curve would look like a cubic except they never touch x = 2 or x = -2
Is this right?

And what about these:
y = 2x / (x^2 -4)
y = (x^2 + 1) / (x^2 - 4)

yeah that's pretty much right...cept for horizontal assymptotes its the limit as x --> infinity..dividing everything by the highest power of x is a way of figuring that out though.

also test points.

have you guys learned about stationary points yet? (where dy/dx = 0)

 

[watatank]

no idea what they would look like. test points to figure it out

also for y intercept you make x=0, x intercept you make y = 0, you did it the wrong way.

for y = 1 / (x^2 - 4)
the horizontal asymptote is y = 0
the vertical asymptotes are x = 2, x = -2
the y intercept is -1/4
the x intercept doesn't exist

for y = 2x / (x^2 -4)
the horizontal asymptote is y = 0
the vertical asymptotes are x = 2, x = -2
the y intercept is 0
the x intercept is 0

y = (x^2 + 1) / (x^2 - 4)
the horizontal asymptote is y = 1
the vertical asymptotes are x = 2, x = -2
the y intercept is -1/4
the x intercept doesnt exist

 

[ronnknee]

Thanks for your replies

for y = 2x / (x^2 -4)
the horizontal asymptote is y = 0
the vertical asymptotes are x = 2, x = -2
the y intercept is 0
the x intercept is 0

Okay, this is where it confuses me. How can there be horizontal asymptote at zero when there is a x intercept at zero? One suggests that it approaches to zero but never touch and the other suggests that it cuts the x axis.

Contradiction

Edit: Actually never mind, I was being narrow minded. I forgot that there is more than one curve

 

[dan.121212]

could someone run through finding horizontal assymptotes again?

also ive seen grpahs which require an assymptote of y=x or something of the like - how would u find that?

thx a heap

 

[dan.121212]

Generally, you must consider the "large x" behaviour of the function.

For example:

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

as x gets big, the effect of the -2 in the numerator and the -4 in the denominator becomes insignificant, so the function will approach x/x, or one. Of course, the numerator is always a little bit bigger than the denominator, so we can see that y will always be a little bit larger than one. in other words, it approaches one from above.

Now, if we consider what happens when x gets large and negative, we can still see that y approaches 1, but this time from below.

If you wish to do this algebraically, then you can re-write

y = (x-2)/(x-4) = (x-4 +2)/(x-4)
=1+ 2/(x-4)

and then it is quite clear that y approaches 1 as x gets large.

thx for replyin Iruka

can you please explain how u did it algebraically, i understand everything above that

thx

EDIT: i think i see it now - so u just re adjusted the way u wrote the equation to have a constant (1)