Rational Functions and Asymptotes

[shanynanigans]

This is for my Pre-Calc class.

Problem: Identify any vertical and horizontal asymptotes.
17. f(x)= x[sup:41pb2664]2[/sup:41pb2664]-25/x[sup:41pb2664]2[/sup:41pb2664]+5x

I factored it to get (x+5)(x+1) And got that the horizontal asymptote was 1, but I don't understand how to get the vertical asymptote.

 

[galactus]

The V.A. is found by setting the denominator equal to 0.

Factor the denominator: \(\displaystyle x(x+5)\)

What values of x make this equal to 0?.

Note that the x+5 cancels with the x+5 in the numerator. Here is a 'hole', not a V.A.

There is only one value that gives a V.A. See it?.

\(\displaystyle \frac{x^{2}-25}{x^{2}+5x}=\frac{(x-5)(\not{x}\not{+}\not{5})}{x(\not{x}\not{+}\not{5})}\)

 

[lookagain]

.

Problem: Identify any vertical and horizontal asymptotes.
17. f(x)= x[sup:3fk0k2rn]2[/sup:3fk0k2rn]-25/x[sup:3fk0k2rn]2[/sup:3fk0k2rn]+5x

shanynanigans,

What you posted is \(\displaystyle x^2 - \frac{25}{x^2} + 5x.\)


Instead, if you're going to type it horizontally, you must have grouping symbols as in:

(x[sup:3fk0k2rn]2[/sup:3fk0k2rn] - 25)/(x[sup:3fk0k2rn]2[/sup:3fk0k2rn] + 5x) or \(\displaystyle (x^2 - 25)/(x^2 + 5x)\)