Aznshorty67

${\mbox{Find all asymptotes and extrema (by hand). Sketch a graph.}}$

$y={\frac {x^{2}}{x^{2}-4x+3}}$

${\mbox{First, see where the function is undefined to find the asymptotes. Set the denominator to zero.}}$

$x^{2}-4x+3=0$

${\mbox{Solve for }}x.$

$(x-3)(x-1)=0$

$(x-3)=0\ \ \ \ \ (x-1)=0$

$x=3\ \ \ \ \ \ \ \ \ \ \ \ \ x=1$

${\mbox{Vertical asymptotes are at }}x=1{\mbox{ and }}x=3$

${\mbox{Since the highest powers of both the numerator and the }}$

${\mbox{denominator are the same, you can divide the coefficients to get the horizontal asymptotes.}}$

$y={\frac {1x^{2}}{1x^{2}-4x+3}}$

$y={\frac {1}{1}}$

${\mbox{The horizontal asymptote is at }}y=1$

${\mbox{To find the extrema, first find the derivative using the quotient rule.}}$

${\frac {dy}{dx}}={\frac {(x^{2}-4x+3)(2x)-(2x-4)(x^{2})}{(x^{2}-4x+3)^{2}}}$

${\mbox{Then find the critical numbers by finding where the derivative is zero or undefined.}}$

${\mbox{First set the numerator to zero to find where the derivative is zero.}}$$(x^{2}-4x+3)(2x)-(2x-4)(x^{2})=0$

${\mbox{Distribute and simplify.}}$

$2x^{3}-8x^{2}+6x-2x^{3}+4x^{2}=0$

$-4x^{2}+6x=0$

${\mbox{Solve for }}x.$

$x(-4x+6)$

$x=0\ \ \ \ \ x={\frac {3}{2}}$

${\mbox{Now set the denominator to zero to find where the derivative is undefined.}}$

$(x^{2}-4x+3)^{2}=0$

$(x-1)(x-3)=0$

$x=1\ \ \ \ \ \ x=3$

${\mbox{The critical numbers are}}$

$x=0,1,{\frac {3}{2}},3$

${\mbox{Test if the critical numbers are extrema using the first derivitave test. }}$

${\mbox{Interval}}$	$(-\infty ,0)$	$(0,1)$	$(1,{\frac {3}{2}})$	$({\frac {3}{2}},3)$	$(3,\infty )$
${\mbox{Test value}}$	$-1$	${\frac {1}{2}}$	${\frac {5}{4}}$	$2$	$4$
${\mbox{Sign of }}f'(x)$	${\mbox{Negative}}$	${\mbox{Positive}}$	${\mbox{Positive}}$	${\mbox{Negative}}$	${\mbox{Negative}}$
${\mbox{Conclusion}}$	$f(x)\ {\mbox{is decreasing}}$	$f(x)\ {\mbox{is increasing}}$	$f(x)\ {\mbox{is increasing}}$	$f(x)\ {\mbox{is decreasing}}$	$f(x)\ {\mbox{is decreasing}}$

$x=0{\mbox{ is a local minimum because }}f'(x){\mbox{ changes from negative to positive at }}x=0$

$x={\frac {3}{2}}{\mbox{ is a local maximum because }}f'(x){\mbox{ changes from positive to negative at }}x={\frac {3}{2}}$

$\mathbf {KenKuo,SergioLeal,PatrickLu,StevenMaheshwary}$

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