009A Sample Final A, Problem 6

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6. Find the vertical and horizontal asymptotes of the function  

Foundations:  
Vertical asymptotes occur whenever the denominator of a rational function goes to zero, and  it doesn't cancel from the numerator.
On the other hand, horizontal asymptotes represent the limit as goes to either positive or negative infinity.

 Solution:

Vertical Asymptotes:  
Setting the denominator to zero, we have
     Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 0=10x-20=10(x-2),}
which has a root at This is our vertical asymptote.
Horizontal Asymptotes:  
More work is required here. Since we need to find the limits at , we can multiply our by

    

This expression is equal to for positive values of , and is equal to for negative values of . Since multiplying by an expression equal to doesn't change the limit, we will add a negative sign to our fraction when considering the limit as goes to . Thus,

     Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{x\rightarrow \pm \infty }{\frac {\sqrt {4x^{2}+3}}{10x-20}}\,\,\cdot \,\,\pm {\frac {\sqrt {\frac {1}{x^{2}}}}{\,\,\,{\frac {1}{x}}}}=\lim _{x\rightarrow \pm \infty }\pm {\frac {\sqrt {{\frac {4x^{2}}{x^{2}}}+{\frac {3}{x^{2}}}}}{{\frac {10x}{x}}-{\frac {20}{x}}}}=\lim _{x\rightarrow \pm \infty }\pm {\frac {\sqrt {4+{\frac {3}{x^{2}}}}}{10-{\frac {20}{x}}}}=\pm {\frac {2}{10}}=\pm {\frac {1}{5}}}

Thus, we have a horizontal asymptote at Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y=-1/5} on the left (as goes to ), and a horizontal asymptote at on the right (as goes to ).

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