Answer
$$
y=\frac{2x^{2}+3}{x^{2}-1}
$$
The values $x= \pm 1$ makes the denominator 0, but not the numerator, so the lines with equations
$$
x=1 , \, x=-1
$$
as vertical asymptotic.
To find a horizontal asymptotic, let $x$ get larger and larger, so that $2x^{2}+3 \approx 2x^{2}$ because the 3 is very small compared with $2x^{2}$. Similarly, for $ x$ very large $x^{2}-1 \approx x^{2}$ . Therefore,
$$
y=\frac{2x^{2}+3}{x^{2}-1} \approx \frac{2x^{2}}{x^{2}} \approx 2.
$$
This means that the line $y=2 $ is a horizontal asymptotic.
When $x=0 $ the y-intercept is -3
This is graph $B$.
Work Step by Step
$$
y=\frac{2x^{2}+3}{x^{2}-1}
$$
The values $x= \pm 1$ makes the denominator 0, but not the numerator, so the lines with equations
$$
x=1 , \, x=-1
$$
as vertical asymptotic.
To find a horizontal asymptotic, let $x$ get larger and larger, so that $2x^{2}+3 \approx 2x^{2}$ because the 3 is very small compared with $2x^{2}$. Similarly, for $ x$ very large $x^{2}-1 \approx x^{2}$ . Therefore,
$$
y=\frac{2x^{2}+3}{x^{2}-1} \approx \frac{2x^{2}}{x^{2}} \approx 2.
$$
This means that the line $y=2 $ is a horizontal asymptotic.
When $x=0 $ the y-intercept is -3
This is graph $B$.
