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Find the vertical asymptotes and removable discontinuities of the graph of f ( x ) = x 2 25 x 3 6 x 2 + 5 x .

Removable discontinuity at x = 5. Vertical asymptotes: x = 0 ,   x = 1.

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Identifying horizontal asymptotes of rational functions

While vertical asymptotes describe the behavior of a graph as the output gets very large or very small, horizontal asymptotes help describe the behavior of a graph as the input gets very large or very small. Recall that a polynomial’s end behavior will mirror that of the leading term. Likewise, a rational function’s end behavior will mirror that of the ratio of the function that is the ratio of the leading terms.

There are three distinct outcomes when checking for horizontal asymptotes:

Case 1: If the degree of the denominator>degree of the numerator, there is a horizontal asymptote    at y = 0.

Example:  f ( x ) = 4 x + 2 x 2 + 4 x 5

In this case, the end behavior is f ( x ) 4 x x 2 = 4 x . This tells us that, as the inputs increase or decrease without bound, this function will behave similarly to the function g ( x ) = 4 x , and the outputs will approach zero, resulting in a horizontal asymptote at y = 0. See [link] . Note that this graph crosses the horizontal asymptote.

Graph of f(x)=(4x+2)/(x^2+4x-5) with its vertical asymptotes at x=-5 and x=1 and its horizontal asymptote at y=0.
Horizontal asymptote y = 0 when f ( x ) = p ( x ) q ( x ) , q ( x ) 0 where degree of p < degree of  q .

Case 2: If the degree of the denominator<degree of the numerator by one, we get a slant asymptote.

Example:  f ( x ) = 3 x 2 2 x + 1 x 1

In this case, the end behavior is f ( x ) 3 x 2 x = 3 x . This tells us that as the inputs increase or decrease without bound, this function will behave similarly to the function g ( x ) = 3 x . As the inputs grow large, the outputs will grow and not level off, so this graph has no horizontal asymptote. However, the graph of g ( x ) = 3 x looks like a diagonal line, and since f will behave similarly to g , it will approach a line close to y = 3 x . This line is a slant asymptote.

To find the equation of the slant asymptote, divide 3 x 2 2 x + 1 x 1 . The quotient is 3 x + 1 , and the remainder is 2. The slant asymptote is the graph of the line g ( x ) = 3 x + 1. See [link] .

Graph of f(x)=(3x^2-2x+1)/(x-1) with its vertical asymptote at x=1 and a slant asymptote aty=3x+1.
Slant asymptote when f ( x ) = p ( x ) q ( x ) , q ( x ) 0 where degree of p > degree of  q by 1 .

Case 3: If the degree of the denominator = degree of the numerator, there is a horizontal asymptote at y = a n b n , where a n and b n are the leading coefficients of p ( x ) and q ( x ) for f ( x ) = p ( x ) q ( x ) , q ( x ) 0.

Example:  f ( x ) = 3 x 2 + 2 x 2 + 4 x 5

In this case, the end behavior is f ( x ) 3 x 2 x 2 = 3. This tells us that as the inputs grow large, this function will behave like the function g ( x ) = 3 , which is a horizontal line. As x ± , f ( x ) 3 , resulting in a horizontal asymptote at y = 3. See [link] . Note that this graph crosses the horizontal asymptote.

Graph of f(x)=(3x^2+2)/(x^2+4x-5) with its vertical asymptotes at x=-5 and x=1 and its horizontal asymptote at y=3.
Horizontal asymptote when f ( x ) = p ( x ) q ( x ) , q ( x ) 0 where degree of  p = degree of  q .

Notice that, while the graph of a rational function will never cross a vertical asymptote    , the graph may or may not cross a horizontal or slant asymptote. Also, although the graph of a rational function may have many vertical asymptotes, the graph will have at most one horizontal (or slant) asymptote.

Questions & Answers

If c is the cost function for a particular product, find the marginal cost functions and their values at x=10 a. c(x) = 800+ 0.04x + 0.0002x² b. c(x) = 250 + 100x + 0.001x²
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9(y+8)-27 is 9y+45. Why can't you reduce that to y+5? I know that's wrong but can't explain why
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when you reduce an equation to its simplest terms, you can't change the value of the equation. reducing it to y + 5 is equivalent to dividing it by 9 which changes the value. you can multiply it by 1 or 9/9 which would give 9(y + 5). multiplying it by one does not change the value.
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1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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