Cubic/cubic ->intercepts and asymptote(s)

[slain4ever]

Homework Statement

Draw a careful sketch graph, showing and labelling all intercepts and asymptotes of:

y = (x^3 - 3x + 2) / (x^3 - 3x^2 + 4)

The Attempt at a Solution

Well the y intercept is simple enough,
let x = 0
which leaves y = 2/4 = 1/2

Im not so sure about the x intercepts,

the asymptote I am guessing is going to be when x^3 - 3x^2 + 4 = 0
therefore asymptote at y = x^3 - 3x^2 + 4
right?

 

[Mark44]

Homework Statement

Draw a careful sketch graph, showing and labelling all intercepts and asymptotes of:

y = (x^3 - 3x + 2) / (x^3 - 3x^2 + 4)

The Attempt at a Solution

Well the y intercept is simple enough,
let x = 0
which leaves y = 2/4 = 1/2

Im not so sure about the x intercepts,

The x-intercepts occur when the numerator is zero; IOW, when x³ - 3x + 2 = 0.

the asymptote I am guessing is going to be when x^3 - 3x^2 + 4 = 0

Yes, but there might be as many as three of them.

therefore asymptote at y = x^3 - 3x^2 + 4
right?

No. The asymptotes are vertical lines x = A, x = B, and x = C, where A, B, and C are the solutions to the equation x³ - 3x² + 4 = 0.

I don't know for certain if there are three of them.

 

[eumyang]

OP: when you say asymptotes, do you mean vertical asymptotes only? Because I believe there is a horizontal asymptote as well. Do you know how to find it?

 

[Mentallic]

And one more thing, if both cubics have a common factor (same root) then it is neither an asymptote nor a root at that x value.

 

[slain4ever]

i think we need all asymptotes, in class we have done angular asymptotes as well

 

[Mentallic]

i think we need all asymptotes, in class we have done angular asymptotes as well

Right, so can you find the horizontal asymptote then?

 

[slain4ever]

ok here's what i have as of now

y intercept is when x = 0; so y = 1/2
x .cepts are at x^3 - 3x + 2 = 0; so x = -2 and x = 1

from the graph I've drawn on my calc this looks about right

asymptotes at x^3 -3x^2 +4 = 0
which gives x = -1 and x = 2

but i have no idea how this relates to my graph because there is definitely no asymptotes at those lines.

so to answer your question; no, i have no idea how to find the asmymptotes

 

[Mentallic]

Ok so the vertical asymptotes are at x=-1,2 the x-intercepts are at x=-2,1 and to find the horizontal asymptotes, think about what happens as x approaches a really big positive or negative number, that is, evaluate \lim_{x\to \pm \infty}f(x)

 

[slain4ever]

i think as x got to infin y would be 1
and as it got -infin y would be 1

 

[slain4ever]

oh i just realized i forgot to multiply the denominator x squared term by 3, so i retract my statement about there being no asymptotes at x=2 and x =-1 on the calculator graph

but i don't see why you all think there is a horizontal asymptote because at x=1 y=0 and at x<2 y>0

 

[slain4ever]

ive looked at the graph and realized that if you ignore the middle part x=-1 to 2 (or -1<x<2) there will in fact be an asymptote at y=1, is it still an asymptote though? i was taught nothing crosses an asymptote, and in this equation there are two values for y = 1and if it is counted as an asymptote, does "label all intercepts" include intercepts with the asymptote?

 

[eumyang]

I was taught that a horizontal asymptote is an end behavior asymptote - the graph of a function with a horizontal asymptote will approach some y-value at the "ends" of the graph. (Now, some graphs will have a horizontal asymptote when you look at one "end" of the graph but not at the other "end" -- look at exponential functions, for instance.) So it is okay if a graph crosses its horizontal asymptote in the middle of the graph. The middle of the graph is not the left "end" or right "end" of the graph.

 

[slain4ever]

allright, so last question is, does where the graph cross the horizontal asymptote count as an 'intercept', and therefore should i include the values on my sketch?

 

[HallsofIvy]

No, it does not. The only "intercepts" here are the x and y intercepts- where the graph crosses the x and y axes.

The y intercepts are at (x, 0) where x satisifies x^3 - 3x + 2= 0. It is easy to see, by inspection, that x= 1 is one root.

They vertical asymptotes are where the denominator is 0: x^3 - 3x^2 + 4 and it is easy to see that x= -1 is a root.

 

[Mentallic]

allright, so last question is, does where the graph cross the horizontal asymptote count as an 'intercept', and therefore should i include the values on my sketch?

I would only add that extra info in if there are a lot of marks for that question and it doesn't take too much time to do. In this case, it'll be relatively easy to find the x value of where the curve cuts the line y=1 since the solution is a root of a quadratic.

1=\frac{x^3-3x+2}{x^3-3x^2+4}

x^3-3x^2+4=x^3-3x+2

3x^2-3x-2=0

However it isn't necessary.