Graphing Rational Functions

[carebear]

I have been asked to sketch y = x²/(x² + 2x - 8)

According to the lesson notes it has a horizontal asymptote at y = 1 because the degree of the numerator is equal to the degree of the denominator ( said to divide the leading coefficients). However, is it really a horizontal asymptote if at least 1 part of the graph exists when y = 1. For example....in this question, when x = 4, y = 1.

When I look at the left side of the graph alone, it definitely has a horizontal asymptote of y = 1.

So can I say the graph has a horizontal asymptote at y= 1?

 

[Deleted member 4993]

I have been asked to sketch y = x²/(x² + 2x - 8)

According to the lesson notes it has a horizontal asymptote at y = 1 because the degree of the numerator is equal to the degree of the denominator ( said to divide the leading coefficients). However, is it really a horizontal asymptote if at least 1 part of the graph exists when y = 1. For example....in this question, when x = 4, y = 1.

When I look at the left side of the graph alone, it definitely has a horizontal asymptote of y = 1.

So can I say the graph has a horizontal asymptote at y= 1?

Yes - the graph can even intersect horizontal assymptote.

The graph of a function cannot intersect its vertical assymptote.

 

[availchet]

here is your graph