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In the function f(x) = x² - c² / (x + a)(x - b)(x + c), where a, b, and c are positive real numbers, which of the following is true?

A) The y-intercept is 0, x-intercept is at a, no discontinuities, and there are no asymptotes.
B) The y-intercept is -c², x-intercept is at -a, no discontinuities, and there is a vertical asymptote at -b.
C) The y-intercept is -c², x-intercept is at b, no discontinuities, and there is a horizontal asymptote at a.
D) The y-intercept is c², x-intercept is at -c, there are discontinuities at a and b, and there is a vertical asymptote at -a.

User Galen Long
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1 Answer

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Final answer:

The most accurate option is B: The y-intercept is -c², x-intercept is at -a, no discontinuities, and there is a vertical asymptote at -b.

Step-by-step explanation:

The function f(x) = x² - c² / (x + a)(x - b)(x + c) represents a rational function. Let's analyze each option to determine which is true:

Option A: The y-intercept is 0, x-intercept is at a, no discontinuities, and there are no asymptotes. This option is incorrect because the y-intercept is -c², not 0, and there may be discontinuities depending on the values of a, b, and c.

Option B: The y-intercept is -c², x-intercept is at -a, no discontinuities, and there is a vertical asymptote at -b. This option is partially correct. The y-intercept and x-intercept are correct, but there may be other asymptotes depending on the values of a, b, and c.

Option C: The y-intercept is -c², x-intercept is at b, no discontinuities, and there is a horizontal asymptote at a. This option is incorrect because the y-intercept is -c², not at -c², and there may be other asymptotes depending on the values of a, b, and c.

Option D: The y-intercept is c², x-intercept is at -c, there are discontinuities at a and b, and there is a vertical asymptote at -a. This option is false because the y-intercept is -c², not c².

Based on the above analysis, Option B is the most accurate: The y-intercept is -c², x-intercept is at -a, no discontinuities, and there is a vertical asymptote at -b.

User Steve Hawkins
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