Why Does My Graph Cross the Horizontal Asymptote y=1?

[Immutef]

Homework Statement

Find the horizontal asymptote(if there is one) using the rule for determining the horizontal asymptote of a rational function for (x^2+x-12)/ (x^2 -4)

Homework Equations

The Attempt at a Solution

the degree of the numerator and denominator are both 2.

Y=(An)/(Bn)
Y=1/1
Y=1

When I do the math, the horizontal asymptote is the line y=1.

However when I graph this equation on a TI- 84 plus graphing calculator, if i use the trace, or table functions, the part of the graph that does not appear to cross the line y=1, does. Why is this?

 

[Mark44]

Homework Statement

Find the horizontal asymptote(if there is one) using the rule for determining the horizontal asymptote of a rational function for (x^2+x-12)/ (x^2 -4)

Homework Equations

The Attempt at a Solution

the degree of the numerator and denominator are both 1.

No, the degree of the numerator and denominator is 2. How did you get 1?

Y=(An)/(Bn)
Y=1/1
Y=1

When I do the math, the horizontal asymptote is the line y=1.

However when I graph this equation on a TI- 84 plus graphing calculator, if i use the trace, or table functions, the part of the graph that does not appear to cross the line y=1, does. Why is this?

The whole idea about a horizontal asymptote is to describe behavior of the function for very large x or very negative x. For very large (or very negative) values, the graph of the function won't cross the asymptote. For values of x that are relatively close to 0, the graph can cross the asymptote.

 

[Immutef]

No, the degree of the numerator and denominator is 2. How did you get 1?
The whole idea about a horizontal asymptote is to describe behavior of the function for very large x or very negative x. For very large (or very negative) values, the graph of the function won't cross the asymptote. For values of x that are relatively close to 0, the graph can cross the asymptote.

Sorry, yes the degree is 2, I typed the wrong number. Thank you for the quick response.

* I will correct that in the original posting, as it was a repeated typo

 

[Mark44]

Now this part is wrong -

Y=(An)/(Bn)
Y=2/2
Y=1

There are two things going on here: (1) the degrees of numerator and denominator, (2) the coefficients of the leading terms in the numerator and denominator.

In a rational function, when deg(numerator) = deg(denominator), the equation of the horizontal asymptote is y = a_n/b_n. Here, a_n is the coefficient of the highest degree term in the numerator, and b_n is the coefficient of the highest degree term in the denominator.

For your problem a_n = b_n = 1.