91Ó°ÊÓ

Vertical asymptotes occur where the denominator of a rational function equals zero and the numerator is non-zero. This is because dividing by zero is undefined in mathematics. For the functions provided:

- For \(f(x) = \frac{x+a}{x+b}\), the vertical asymptote is found by setting the denominator to zero: \(x+b=0 \)', therefore, \(x=-b\).
- For \(g(x)=\frac{x+a}{(x+b)(x+c)} \), vertical asymptotes are found by setting each factor in the denominator to zero: \(x=-b\) and \(x=-c\).
- Similarly, for \(h(x)=\frac{(x+a)(x+c)}{(x+b)(x+c)} \), there is a vertical asymptote where \(x=-b\). However, the \(x=-c\) does not create an asymptote because it cancels out with the term in the numerator.

In summary, rational functions have vertical asymptotes wherever the denominator equals zero and the numerator does not also equal zero at that point.

Horizontal asymptotes describe the behavior of rational functions as \(x \rightarrow \infty\) or \(x \rightarrow -\infty\). They depend on the degrees of the polynomials in the numerator and denominator.

- For \(f(x)=\frac{x+a}{x+b} \), the degrees of the numerator and the denominator are both 1, so the horizontal asymptote is determined by the ratio of the leading coefficients: 1. The horizontal asymptote is therefore \(y=1\).

- For \(g(x)=\frac{x+a}{(x+b)(x+c)} \), the degree of the denominator (2) is greater than the degree of the numerator (1). Hence, the horizontal asymptote is at \(y=0\).

- For \(h(x)=\frac{(x+a)(x+c)}{(x+b)(x+c)} \), after simplifying to \( \frac{x+a}{x+b}\), we see that the degrees of the numerator and denominator are both equal to 1. Therefore, the horizontal asymptote is \(y=1 \).

In summary, the horizontal asymptote for these functions can be found by comparing the degrees of the polynomials or simply observing the simplified form.

Simplifying rational functions involves cancelling out common factors in the numerator and denominator and making the expression as simple as possible.

- For \(f(x)=\frac{x+a}{x+b}\), the function is already simplified because there are no common factors between the numerator and the denominator.
- For \(g(x)=\frac{x+a}{(x+b)(x+c)}\), this function is already in its simplest form for the same reason; there's no common factor.
- For \(h(x)=\frac{(x+a)(x+c)}{(x+b)(x+c)}\), the term \(x+c\) is present both in the numerator and the denominator, so they can be cancelled out to simplify the function to \( \frac{x+a}{x+b}\).

When simplifying, always check for factors in the numerator and denominator that can be cancelled. This will make analyzing and graphing the functions easier.