MHB What is the correct slant asymptote for this function and why is it significant?

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The discussion clarifies the concept of slant asymptotes in relation to the function given. It emphasizes that while the function approaches the slant asymptote, it never actually crosses it due to the presence of a non-zero remainder in the long division. The calculation shows that the function behaves like -3x + 2 as it approaches infinity, confirming that this line is indeed the slant asymptote. The confusion around the graphical representation of asymptotes is addressed, highlighting the importance of understanding the behavior of the function at extreme values. Overall, the significance of the slant asymptote lies in its role in describing the end behavior of the function.
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View attachment 819Why is the slant asymptote pictured here correct for this function? I was under the impression an asymptote was never crossed by the function. I get that the dividend gives the equation for the asymptote for a non zero remainder, but seeing this graphically is a bit confusing. Thanks! (EDIT: NVM, I just answered my own question:p) Thanks anyways though!
 

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If you long divide, you'll see that [math]\displaystyle \frac{-6x^3 + 4x^2 - 1}{2x^2 + 1} \equiv -3x + 2 + \frac{x}{2x^2 + 1} = -3x + 2 + \frac{1}{2x + \frac{1}{x}} [/math].

You can see that the denominator easily overpowers the numerator, and so gets closer to 0, which means the entire function eventually works like [math]\displaystyle -3x + 2[/math]. However, since there is always that tiny bit added, it never actually will be [math]\displaystyle -3x + 2[/math], and so [math]\displaystyle y = -3x + 2[/math] must be an asymptote :)
 
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