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A horizontal asymptote gives us a sense of where a function will level out as it heads towards positive or negative infinity. Here's how you find one:

• Compare the degrees (the highest power of the variable) of the numerator and denominator of the function.
• If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is the x-axis, or in mathematical terms, \( y = 0 \).
• If the degrees are the same, the horizontal asymptote is given by the ratio of their leading coefficients. So if each has a leading coefficient of 1, \( y = 1 \).

Let's break it down with examples:- **For** \( f(x)=\frac{-x-3}{x^2-5x-14} \), the degree is higher in the denominator, resulting in a horizontal asymptote at \( y = 0 \).- **For** \( r(x)=\frac{x^2+x-6}{x^2-16} \), both the numerator and denominator have the same degree, which places the horizontal asymptote at \( y = 1 \).These asymptotes help us visualize how the graph of the function behaves as it stretches towards infinity.

The oblique or slant asymptote might seem tricky at first, but it's all about the degree of the functions:

• They appear when the degree of the numerator is exactly one higher than the degree of the denominator.
• To find an oblique asymptote, perform polynomial long division on the functions involved.

For our problem, let's see how it works:- **For** \( h(x)=\frac{x^3-1}{x^2+4} \), we find the numerator's degree is one more than the denominator's degree. By dividing \( x^3 - 1 \) by \( x^2 + 4 \), we get an oblique asymptote at \( y = x \).The graph not only tells us about a function's direction as it goes to infinity, but for slanting lines, it offers a linear boundary the function will approach but never reach.

Vertical asymptotes are found by identifying points where the function's denominator equals zero while the numerator does not:

• These lines occur when a division by zero happens that doesn't get "cancelled out" by the numerator.
• They often represent values of \( x \) that produce outputs approaching infinity as we approach these points.

For our functions, let's examine:- **For** \( f(x)=\frac{-x-3}{x^2-5x-14} \), solving \( x^2 - 5x - 14 = 0 \) gives roots at \( x = -2 \) and \( x = 7 \), creating vertical asymptotes there.- **For** \( r(x)=\frac{x^2+x-6}{x^2-16} \), \( x^2 - 16 = 0 \) produces vertical asymptotes at \( x = -4 \) and \( x = 4 \).However, for \( g(x)=\frac{x-x^3}{x-3} \), the numerator and denominator reach zero simultaneously at \( x = 3 \), eliminating vertical asymptotes due to simplification. Understanding these points allows for accurate depiction of function behavior rather than assumptions of undefined values.