Difference between revisions of "009A Sample Final A, Problem 6"

Revision as of 09:38, 25 March 2015

6. Find the vertical and horizontal asymptotes of the function $f(x)={\frac {\sqrt {4x^{2}+3}}{10x-20}}.$

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 x goes to either positive or negative infinity.

Vertical Asymptotes:
Setting the denominator to zero, we have
$0=10x-20=10(x-2),$
which has a root at x = 2. This is our vertical asymptote.

Horizontal Asymptotes:
More work is required here. Since we need to find the limits at $\pm \infty$ , we can multiply our f(x) by
${\frac {\sqrt {\frac {1}{x^{2}}}}{\,\,\,{\frac {1}{x}}}}.$
This expression is equal to 1 for positive values of x, and is equal to -1 for negative values of x. Since multiplying f(x) by an expression equal to 1 doesn't change the limit, we will add a negative sign to it when considering the limit as x goes to $-\infty$ . Thus,
$\lim _{x\rightarrow \pm \infty }{\frac {\sqrt {4x^{2}+3}}{10x-20}}\,\,\cdot \,\,\pm {\frac {\sqrt {\frac {1}{x^{2}}}}{\,\,\,{\frac {1}{x}}}}=\lim _{x\rightarrow \pm \infty }\pm {\frac {\sqrt {{\frac {4x^{2}}{x^{2}}}+{\frac {3}{x^{2}}}}}{{\frac {10x}{x}}-{\frac {20}{x}}}}=\lim _{x\rightarrow \pm \infty }\pm {\frac {\sqrt {4+{\frac {3}{x^{2}}}}}{10-{\frac {20}{x}}}}=\pm {\frac {2}{10}}=\pm {\frac {1}{5}}$
Thus, we have a horizontal asymptote at y = -1/5 on the left (as x goes to $-\infty$ ), and a horizontal asymptote at y = 1/5 as x goes to $+\infty$ ).

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 |<br>&nbsp;&nbsp;&nbsp;&nbsp;  <math>\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}</math>
 |-
-|<br>Thus, we have a horizontal asymptote at ''y''=-1/5 on the left (as ''x'' goes to <math style="vertical-align: -5%;">-\infty</math>), and a horizontal asymptote at ''y''=1/5 as ''x'' goes to  <math style="vertical-align: -5%;">+\infty</math>).
+|<br>Thus, we have a horizontal asymptote at ''y'' = -1/5 on the left (as ''x'' goes to <math style="vertical-align: -5%;">-\infty</math>), and a horizontal asymptote at ''y'' = 1/5 as ''x'' goes to  <math style="vertical-align: -5%;">+\infty</math>).
 |}
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