Try explaining how you worked the problem and how you did it
Answer: f(x) will have vertical asymptotes at x=-2 and x=2 and horizontal asymptote at y=3.
Step-by-step explanation:
Given function: ![f(x)=\dfrac{3x^2}{x^2-4}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cdfrac%7B3x%5E2%7D%7Bx%5E2-4%7D)
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://tex.z-dn.net/?f=x%5E2-4%3D0%5CRightarrow%5C%20x%5E2%3D4%5CRightarrow%5C%20x%3D%5Cpm2)
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=\dfrac{\text{Coefficient of }x^2\text{ in numerator}}{\text{Coefficient of }x^2\text{ in denominator}}](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B%5Ctext%7BCoefficient%20of%20%7Dx%5E2%5Ctext%7B%20in%20numerator%7D%7D%7B%5Ctext%7BCoefficient%20of%20%7Dx%5E2%5Ctext%7B%20in%20denominator%7D%7D)
i.e. ![y=\dfrac{3}{1}=3](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B3%7D%7B1%7D%3D3)
Hence, f(x) will have horizontal asymptote at y=3.
-5.14? Not sure if it's correct, but it's my best guess.
THIS THE CORRECT ANSWER :land is primary factor of production
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