Find all asymptotes and extrema (by hand). Sketch a graph. {\displaystyle {\mbox{Find all asymptotes and extrema (by hand). Sketch a graph.}}}
y = x 2 x 2 − 4 x + 3 {\displaystyle y={\frac {x^{2}}{x^{2}-4x+3}}}
First, see where the function is undefined to find the asymptotes. Set the denominator to zero. {\displaystyle {\mbox{First, see where the function is undefined to find the asymptotes. Set the denominator to zero.}}}
x 2 − 4 x + 3 = 0 {\displaystyle x^{2}-4x+3=0}
Solve for x . {\displaystyle {\mbox{Solve for }}x.}
( x − 3 ) ( x − 1 ) = 0 {\displaystyle (x-3)(x-1)=0}
( x − 3 ) = 0 ( x − 1 ) = 0 {\displaystyle (x-3)=0\ \ \ \ \ (x-1)=0}
x = 3 x = 1 {\displaystyle x=3\ \ \ \ \ \ \ \ \ \ \ \ \ x=1}
Vertical asymptotes are at x = 1 and x = 3 {\displaystyle {\mbox{Vertical asymptotes are at }}x=1{\mbox{ and }}x=3}
Since the highest powers of both the numerator and the {\displaystyle {\mbox{Since the highest powers of both the numerator and the }}}
denominator are the same, you can divide the coefficients to get the horizontal asymptotes. {\displaystyle {\mbox{denominator are the same, you can divide the coefficients to get the horizontal asymptotes.}}}
y = 1 x 2 1 x 2 − 4 x + 3 {\displaystyle y={\frac {1x^{2}}{1x^{2}-4x+3}}}
y = 1 1 {\displaystyle y={\frac {1}{1}}}
The horizontal asymptote is at y = 1 {\displaystyle {\mbox{The horizontal asymptote is at }}y=1}
To find the extrema, first find the derivative using the quotient rule. {\displaystyle {\mbox{To find the extrema, first find the derivative using the quotient rule.}}}
d y d x = ( x 2 − 4 x + 3 ) ( 2 x ) − ( 2 x − 4 ) ( x 2 ) ( x 2 − 4 x + 3 ) 2 {\displaystyle {\frac {dy}{dx}}={\frac {(x^{2}-4x+3)(2x)-(2x-4)(x^{2})}{(x^{2}-4x+3)^{2}}}}
Then find the critical numbers by finding where the derivative is zero or undefined. {\displaystyle {\mbox{Then find the critical numbers by finding where the derivative is zero or undefined.}}}
First set the numerator to zero to find where the derivative is zero. {\displaystyle {\mbox{First set the numerator to zero to find where the derivative is zero.}}} ( x 2 − 4 x + 3 ) ( 2 x ) − ( 2 x − 4 ) ( x 2 ) = 0 {\displaystyle (x^{2}-4x+3)(2x)-(2x-4)(x^{2})=0}
Distribute and simplify. {\displaystyle {\mbox{Distribute and simplify.}}}
2 x 3 − 8 x 2 + 6 x − 2 x 3 + 4 x 2 = 0 {\displaystyle 2x^{3}-8x^{2}+6x-2x^{3}+4x^{2}=0}
− 4 x 2 + 6 x = 0 {\displaystyle -4x^{2}+6x=0}
x ( − 4 x + 6 ) {\displaystyle x(-4x+6)}
x = 0 x = 3 2 {\displaystyle x=0\ \ \ \ \ x={\frac {3}{2}}}
Now set the denominator to zero to find where the derivative is undefined. {\displaystyle {\mbox{Now set the denominator to zero to find where the derivative is undefined.}}}
( x 2 − 4 x + 3 ) 2 = 0 {\displaystyle (x^{2}-4x+3)^{2}=0}
( x − 1 ) ( x − 3 ) = 0 {\displaystyle (x-1)(x-3)=0}
x = 1 x = 3 {\displaystyle x=1\ \ \ \ \ \ x=3}
The critical numbers are {\displaystyle {\mbox{The critical numbers are}}}
x = 0 , 1 , 3 2 , 3 {\displaystyle x=0,1,{\frac {3}{2}},3}
Test if the critical numbers are extrema using the first derivitave test. {\displaystyle {\mbox{Test if the critical numbers are extrema using the first derivitave test. }}}
x = 0 is a local minimum because f ′ ( x ) changes from negative to positive at x = 0 {\displaystyle x=0{\mbox{ is a local minimum because }}f'(x){\mbox{ changes from negative to positive at }}x=0}
x = 3 2 is a local maximum because f ′ ( x ) changes from positive to negative at x = 3 2 {\displaystyle x={\frac {3}{2}}{\mbox{ is a local maximum because }}f'(x){\mbox{ changes from positive to negative at }}x={\frac {3}{2}}}
K e n K u o , S e r g i o L e a l , P a t r i c k L u , S t e v e n M a h e s h w a r y {\displaystyle \mathbf {KenKuo,SergioLeal,PatrickLu,StevenMaheshwary} }
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