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Find the vertical, horizontal, and slant asymptotes, if any, for f(x)=5x^3-51x^2+77x+100/x^2-11x+24

asked
User Jcopenha
by
8.2k points

2 Answers

0 votes

Answer:

5x+4

Explanation:

edge 2020 2021

answered
User Amr Badawy
by
7.7k points
2 votes

Answer:

Explanation:

Given: f(x)=5x^3-51x^2+77x+100/x^2-11x+24

Please use parentheses to eliminate any ambiguity:

f(x) = (5x^3-51x^2+77x+100) / (x^2 - 11x + 24)

or (better yet):

5x^3-51x^2+77x+100

f(x) = ---------------------------------

x^2 - 11x + 24

The vertical asymptotes here are at the zeros of the denominator:

x^2 - 11x + 24 = 0, This quadratic equation has coefficients a = 1, b = -11 and c = 24. Thus, its roots (zeros) are:

-(-11) ± √( 121 - 4(1)(24) )

x = -------------------------------------

2(1)

or:

11 ± √( 25 )

x = --------------------

2

or: x = 8 and x = 3

The vertical asymptotes are x = 8 and x = 3.

If we attempt to divide x^2 - 11x + 24 into 5x^3 - 51x^2 + 77x + 100, we see that the first term of the quotient is 5x. As x increases or decreases without bound, 5x goes to either ∞ or -∞, so we conclude that there is no horiz. asymptote. Continuing this division results in:

5x + 4 + a fraction

This represents the slant asymptote, y = 5x + 4

answered
User Igonato
by
7.7k points
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