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# Please find all horizontal asymptotes vertical asym. x and y intercepts (if any)?

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

or

x squared minus four over x squared minus two

### 2 Answers

- 1 decade agoFavorite Answer
X and Y intercepts are the easy parts. If you plug zero in for X you get the Y intercepts. So -4/-2, or 2 is your y intercept.

Plugging 0 in for Y will give you your X intercepts. The only way to make this problem equal to zero is to make the numerator equal to zero. So x^2-4 = 0. x^2=4, so x= +/- 2. Those are your y intercepts.

Vertical asymptotes are only where X can not be plugged in. Here, it's where the denominator is zero. So, set it equal to zero, and you get x^2-2=0, or x^2=2, or x=+/- the square root of two.

Horizontal asymptotes are kind of sketchy ground for mathematicians, but a good rule of thumb is to take your biggest powered X's, and make them into a fraction. So, x^2/x^2, which equals 1. That is your horizontal asymptote.

- burnhamLv 44 years ago
a million. vertical asymptotes happen whilst denominator = 0. if asymptote at x = -2/3 then cx + d could desire to be : 3x +2 2. whilst x = 0 then f(x) = -a million/2. => b/d = -a million/2 => b = -a million 3. x-int of a million/4, 0 = (a/4 -a million)/(3/4 + 2) = ( a-4)/(3+8) = (a-4)/11 possibly superb in case you start up with the intercepts first. => a = 4 f(x) = (4x -a million)(3x+2) as x -> infinity f(x) -> (4 - a million/x)/(3 + 2/x) -> 4/3