help with asymptotes please

[Guest]

Help need to understand how to find vertical and horizonal asymptotes.

problem states find all v/a and h/a for

g(x)= 3x / x^4+2x^2+1

problem 2

solve to 3 places

x /x^2+5x-6 less than or equal to 0.5

x+3 is a zero ?

thanks for the help

 

[tkhunny]

Help need to understand how to find vertical and horizonal asymptotes.

problem states find all v/a and h/a for

g(x)= 3x / x^4+2x^2+1

Fix your notation. I presume you mean g(x)= 3x /( x^4+2x^2+1)

Vertical Asymptotes are zeros of the denominator, unles they are also zeros (of the same degree) of the numerator.

Solve: x^4+2x^2+1 = 0, which has no real roots, by inspection. No vertical asymptotes.

Horizontals are easy enough. If N = the Degree of the Numerator and D = the Degree of the Denominator,

D > N ==> Horizontal Asyptote at y = 0
D = N ==> Horizontal Asymptote other than y = 0
D = N - 1 ==> Linear Asymptote, but it isn't horizontal.
D < N - 1 ==> NonLinear Asymptote of some sort.

problem 2

solve to 3 places

x /x^2+5x-6 less than or equal to 0.5

Find critical values and check.

Vertical asymptotes at x^2 + 5x - 6 = 0, or (x+6)(x-1) = 0, or x = -6 and x = 1.

Solve x /x^2+5x-6 = 0.5. I get x = ½(-3 ± √33)

This chops up the x-axis into five sections. Check them.

x < -6
-6 < x < ½(-3 - √33)
½(-3 - √33) < x < 1
1 < x < ½(-3 + √33)
x > ½(-3 + √33)

x+3 is a zero ?

What is this doing in here?

 

[psycho]

I thought you find horizontal asymptotes by setting the problem as a lim-->infinity.

 

[tkhunny]

I thought you find horizontal asymptotes by setting the problem as a lim-->infinity.

1) I don;t know what that means. Do you mean "finding" the limit?

2) "Horizontal"? It will find onlt the ones that ARE horizontal.

3) The limit is the fundamental concept of asymptotes. There are various other methods for certain special cases, Rational Functions, for example.