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zeros and asymptotes

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What are the zeroes and asymptotes of (2x)/(x^2-4) ?

asked Apr 10, 2013 in ALGEBRA 2 by Jose Rodriguez Rookie

2 Answers

0 votes

Asymptotes :

(2x) / (x^2 - 4)

Vertical asymptotes at 2 , -2      x^2 - 4 = 0

Add 4 to each side

x^2 - 4 + 4 = 4

x^2 + 0 = 4

x^2 = 4

Take square root  to each side

x = +2 , -2.

Horizontal asymptote :

Horizontal asymptote  at 2 2x / x = 2

Zeros :

2x = 0

Divide each side by 2

x = 0.

answered Apr 12, 2013 by diane Scholar
edited Apr 12, 2013 by diane
0 votes

To find zeros, set the function equal to 0.

(2x)/(x^2-4) = 0

(2x)/(x+2) (x -2) = 0

2x = 0. x = 0

so the graph crosses the x -axis at 0.

To find the vertical asymptotes, set the denominator equal to 0.

(x + 2 ) = 0 or x - 2 =0

x = +2, -2

We can't have a zero in the denominator, then domain can't have x = –2 or x = 2.

you can draw a graph by avoiding the vertical lines x = –2 and x = 2.

Source: http://answers.yahoo.com/

answered Apr 13, 2013 by Naren Answers Apprentice

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