Answer: f(x) will have vertical asymptotes at x=-2 and x=2 and horizontal asymptote at y=3.
Explanation:
Given function:
![f(x)=(3x^2)/(x^2-4)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/my1nn34guvpkjvwtgbxw0tnuvceb1ukoaf.png)
The vertical asymptote occurs for those values of x which make function indeterminate or denominator 0.
i.e.
![x^2-4=0\Rightarrow\ x^2=4\Rightarrow\ x=\pm2](https://img.qammunity.org/2021/formulas/mathematics/high-school/ryokllrcrogq8cpdqiat43wqahms94d189.png)
Hence, f(x) will have vertical asymptotes at x=-2 and x=2.
To find the horizontal asymptote , we can see that the degree of numerator and denominator is same i.e. 2.
So, the graph will horizontal asymptote at
![y=\frac{\text{Coefficient of }x^2\text{ in numerator}}{\text{Coefficient of }x^2\text{ in denominator}}](https://img.qammunity.org/2021/formulas/mathematics/high-school/dshitffo6nig42dxrxt6kd6ig2o2nasyoy.png)
i.e.
![y=(3)/(1)=3](https://img.qammunity.org/2021/formulas/mathematics/high-school/a379qkhxynds7sbx8o2hngft8xr1is4kxo.png)
Hence, f(x) will have horizontal asymptote at y=3.