g(x)=2 - 3x^2/X^2 - 1 => 2 - 3x^2 divided by x^2 - 1
How to find asymptotes and intercepts?
Thx
if you are asking what the intercepts and asymptotes of $\displaystyle f(x)=\frac{2-3x^2}{x^2-1}$ are you do this
y-asymp take the $\displaystyle \lim_{x \to {\pm\infty}}\frac{2-3x^2}{x^2-1}$
x-asymp..find the values that make the top 0 but not the bottom 0
x-int...set it equal to 0 and solve...remember in a rational function the only part that makes the function 0 is the numerator...so just solve the numerator of 0
y-int...just take $\displaystyle f(0)$
Find the the asymptotes of
$\displaystyle g(x)=\frac{2-3x^2}{x^2-1}$ I think this is what you mean
Please use parenthsis to make it clear
Our rule for horizontal asympototes is:
if the degree of the numerator is bigger than the degree on the denomitor there isn't one
if the degree of the numerator and denominator are equal it is the ratio of the lead coeffeints
if the degree of the denominator is bigger than the numerator it is the line y=0.
we need the 2nd rule so the asymptote is $\displaystyle y=\frac{-3}{1}=-3$
to find vertical asymptotes we set the denominator equal to zero.
$\displaystyle x^2-1=0 \iff (x-1)(x+1)=0,x= \pm 1$
so x=1 and x=-1 are the vertical asymptotes
to find the intercepts you set the opposite variable = to zero
x-int set y=0
$\displaystyle 0=\frac{2-3x^2}{x^2-1} \iff 0=2-3x^2 \iff 3x^2=2 \iff x =\pm \sqrt{\frac{2}{3}}$
y-int set x=0
$\displaystyle y=\frac{2-3(0)^2}{0^2-1}=-2$
I hope this helps