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A rational function is a mathematical function that can be expressed as the quotient of two polynomials. This means a rational function has the form f(x) = P(x)/Q(x), where P(x) and Q(x) are polynomials, and Q(x) must not be the zero polynomial (Q(x) ≠ 0).
An asymptote of a rational function is a straight line that describes the behavior of the function's graph as x or f(x) approaches certain values but never actually reaches them. There are three types of asymptotes: vertical, horizontal, and slant (or oblique) asymptotes.
For the function f(x) = (2x² + 3x – 5) / (x – 1), let's find the asymptotes.
Set the denominator x – 1 = 0, which gives x = 1.
Now, check the value of the numerator P(x) = 2x² + 3x – 5 at x = 1:
P(1) = 2(1)² + 3(1) – 5 = 2 + 3 – 5 = 0.
Since both the numerator and the denominator are 0 at x = 1, we should factor the numerator:
2x² + 3x – 5 = (x – 1)(2x + 5).
So, f(x) = [(x – 1)(2x + 5)] / (x – 1).
For x ≠ 1, f(x) = 2x + 5.
Because the (x-1) factor cancels, there is no vertical asymptote at x = 1. Instead, there is a hole in the graph at x = 1. The y-coordinate of the hole is 2(1) + 5 = 7.
The degree of the numerator (2) is greater than the degree of the denominator (1). Therefore, there is no horizontal asymptote.
Since the degree of the numerator is exactly one greater than the degree of the denominator, we perform polynomial long division of (2x² + 3x – 5) by (x – 1).
The division yields a quotient of 2x + 5 and a remainder of 0.
So, f(x) = 2x + 5, for x ≠ 1.
In this particular case, because the remainder is zero, the function itself simplifies to the line y = 2x + 5 (with a hole at x=1). This line is the graph of the function, rather than a line the function approaches asymptotically.
(For a typical slant asymptote, the polynomial division would result in a linear quotient mx+b and a non-zero remainder R(x), so f(x) = mx + b + R(x)/Q(x), where R(x)/Q(x) approaches 0 as x approaches ±∞. Then, y = mx + b would be the slant asymptote.)