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Plotting a function graph/ (x^2-2x+1)/(2x-1)

Graphing y = (x^2-2x+1)/(2x-1)

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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        2          
       x  - 2*x + 1
f(x) = ------------
         2*x - 1
f(x)=(x2−2x)+12x−1f{\left(x \right)} = \frac{\left(x^{2} - 2 x\right) + 1}{2 x - 1} 
f = (x^2 - 2*x + 1)/(2*x - 1)
The graph of the function
02468-8-6-4-2-1010-1010
The domain of the function
The points at which the function is not precisely defined:
x1=0.5x_{1} = 0.5
The points of intersection with the X-axis coordinate
Graph of the function intersects the axis X at f = 0
so we need to solve the equation:
(x2−2x)+12x−1=0\frac{\left(x^{2} - 2 x\right) + 1}{2 x - 1} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=1x_{1} = 1
Numerical solution
x1=1.00000022767447x_{1} = 1.00000022767447
x2=1.00000023077877x_{2} = 1.00000023077877
x3=1.00000023331806x_{3} = 1.00000023331806
x4=1.00000022816402x_{4} = 1.00000022816402
x5=1.00000022423582x_{5} = 1.00000022423582
x6=1.00000022833613x_{6} = 1.00000022833613
x7=1.00000020922882x_{7} = 1.00000020922882
x8=1.00000023075242x_{8} = 1.00000023075242
x9=1.00000022678069x_{9} = 1.00000022678069
x10=1.00000023116745x_{10} = 1.00000023116745
x11=1.00000024762859x_{11} = 1.00000024762859
x12=1.00000019809044x_{12} = 1.00000019809044
x13=1.00000023181557x_{13} = 1.00000023181557
x14=1.00000023308419x_{14} = 1.00000023308419
x15=1.00000022591025x_{15} = 1.00000022591025
x16=1.00000022566571x_{16} = 1.00000022566571
x17=1.00000023143781x_{17} = 1.00000023143781
x18=1.0000002271276x_{18} = 1.0000002271276
x19=1.00000023172847x_{19} = 1.00000023172847
x20=1.00000022801895x_{20} = 1.00000022801895
x21=1.0000002313768x_{21} = 1.0000002313768
x22=1.00000022303486x_{22} = 1.00000022303486
x23=1.00000023067968x_{23} = 1.00000023067968
x24=1.00000025596508x_{24} = 1.00000025596508
x25=1.0000002283607x_{25} = 1.0000002283607
x26=1.0000002351771x_{26} = 1.0000002351771
x27=1.00000023080628x_{27} = 1.00000023080628
x28=1.00000022468397x_{28} = 1.00000022468397
x29=1.00000022113012x_{29} = 1.00000022113012
x30=1.00000022838429x_{30} = 1.00000022838429
x31=1.00000023121519x_{31} = 1.00000023121519
x32=1.00000022793426x_{32} = 1.00000022793426
x33=1.00000022828377x_{33} = 1.00000022828377
x34=1.00000023238061x_{34} = 1.00000023238061
x35=1.00000023579214x_{35} = 1.00000023579214
x36=1.00000023083504x_{36} = 1.00000023083504
x37=1.00000023269368x_{37} = 1.00000023269368
x38=1.00000057818485x_{38} = 1.00000057818485
x39=1.00000023190988x_{39} = 1.00000023190988
x40=1.00000023072717x_{40} = 1.00000023072717
x41=1.00000022809513x_{41} = 1.00000022809513
x42=1.00000015959982x_{42} = 1.00000015959982
x43=1.00000022788827x_{43} = 1.00000022788827
x44=1.00000022747379x_{44} = 1.00000022747379
x45=1.00000023224625x_{45} = 1.00000023224625
x46=1.00000023157282x_{46} = 1.00000023157282
x47=1.00000023164778x_{47} = 1.00000023164778
x48=1.00000023358492x_{48} = 1.00000023358492
x49=1.00000022797775x_{49} = 1.00000022797775
x50=1.00000022664173x_{50} = 1.00000022664173
x51=1.0000002253858x_{51} = 1.0000002253858
x52=1.00000022805802x_{52} = 1.00000022805802
x53=1.00000023201233x_{53} = 1.00000023201233
x54=1.00000023212401x_{54} = 1.00000023212401
x55=1.00000022220419x_{55} = 1.00000022220419
x56=1.00000022754533x_{56} = 1.00000022754533
x57=1.00000023425021x_{57} = 1.00000023425021
x58=1.00000022690692x_{58} = 1.00000022690692
x59=1.00000023096439x_{59} = 1.00000023096439
x60=1.00000023103923x_{60} = 1.00000023103923
x61=1.00000023107967x_{61} = 1.00000023107967
x62=1.00000022506225x_{62} = 1.00000022506225
x63=1.00000022631702x_{63} = 1.00000022631702
x64=1.0000002273141x_{64} = 1.0000002273141
x65=1.00000023467217x_{65} = 1.00000023467217
x66=1.00000022369645x_{66} = 1.00000022369645
x67=1.00000022761207x_{67} = 1.00000022761207
x68=1.00000022819605x_{68} = 1.00000022819605
x69=1.00000023131958x_{69} = 1.00000023131958
x70=1.000000231503x_{70} = 1.000000231503
x71=1.00000024063093x_{71} = 1.00000024063093
x72=1.00000021968731x_{72} = 1.00000021968731
x73=1.0000002270221x_{73} = 1.0000002270221
x74=1.00000023070294x_{74} = 1.00000023070294
x75=1.00000023086513x_{75} = 1.00000023086513
x76=1.00000023287757x_{76} = 1.00000023287757
x77=1.00000023100085x_{77} = 1.00000023100085
x78=1.00000023112234x_{78} = 1.00000023112234
x79=1.00000023753679x_{79} = 1.00000023753679
x80=1.00000022773296x_{80} = 1.00000022773296
x81=1.00000023089664x_{81} = 1.00000023089664
x82=1.00000022783956x_{82} = 1.00000022783956
x83=1.00000022612572x_{83} = 1.00000022612572
x84=1.00000023092969x_{84} = 1.00000023092969
x85=1.00000022825583x_{85} = 1.00000022825583
x86=1.00000023655771x_{86} = 1.00000023655771
x87=1.00000021453836x_{87} = 1.00000021453836
x88=1.000000226488x_{88} = 1.000000226488
x89=1.00000021764645x_{89} = 1.00000021764645
x90=1.00000022722461x_{90} = 1.00000022722461
x91=1.00000022778789x_{91} = 1.00000022778789
x92=1.00000023252898x_{92} = 1.00000023252898
x93=1.00000022822663x_{93} = 1.00000022822663
x94=1.00000023126581x_{94} = 1.00000023126581
x95=1.00000023389231x_{95} = 1.00000023389231
x96=1.00000022831051x_{96} = 1.00000022831051
x97=1.00000027857349x_{97} = 1.00000027857349
x98=1.00000023883316x_{98} = 1.00000023883316
x99=1.00000022813042x_{99} = 1.00000022813042
x100=1.00000022739692x_{100} = 1.00000022739692
x101=1.00000024329078x_{101} = 1.00000024329078
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to (x^2 - 2*x + 1)/(2*x - 1).
(02−0)+1−1+0⋅2\frac{\left(0^{2} - 0\right) + 1}{-1 + 0 \cdot 2}
The result:
f(0)=−1f{\left(0 \right)} = -1
The point:
(0, -1)
Extrema of the function
In order to find the extrema, we need to solve the equation
ddxf(x)=0\frac{d}{d x} f{\left(x \right)} = 0
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
ddxf(x)=\frac{d}{d x} f{\left(x \right)} =
the first derivative
2x−22x−1−2((x2−2x)+1)(2x−1)2=0\frac{2 x - 2}{2 x - 1} - \frac{2 \left(\left(x^{2} - 2 x\right) + 1\right)}{\left(2 x - 1\right)^{2}} = 0
Solve this equation
The roots of this equation
x1=0x_{1} = 0
x2=1x_{2} = 1
The values of the extrema at the points:
(0, -1)

(1, 0)


Intervals of increase and decrease of the function:
Let's find intervals where the function increases and decreases, as well as minima and maxima of the function, for this let's look how the function behaves itself in the extremas and at the slightest deviation from:
Minima of the function at points:
x1=1x_{1} = 1
Maxima of the function at points:
x1=0x_{1} = 0
Decreasing at intervals
(−∞,0]∪[1,∞)\left(-\infty, 0\right] \cup \left[1, \infty\right)
Increasing at intervals
[0,1]\left[0, 1\right]
Inflection points
Let's find the inflection points, we'll need to solve the equation for this
d2dx2f(x)=0\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
d2dx2f(x)=\frac{d^{2}}{d x^{2}} f{\left(x \right)} =
the second derivative
2(−4(x−1)2x−1+1+4(x2−2x+1)(2x−1)2)2x−1=0\frac{2 \left(- \frac{4 \left(x - 1\right)}{2 x - 1} + 1 + \frac{4 \left(x^{2} - 2 x + 1\right)}{\left(2 x - 1\right)^{2}}\right)}{2 x - 1} = 0
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
x1=0.5x_{1} = 0.5
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞((x2−2x)+12x−1)=−∞\lim_{x \to -\infty}\left(\frac{\left(x^{2} - 2 x\right) + 1}{2 x - 1}\right) = -\infty
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
lim⁡x→∞((x2−2x)+12x−1)=∞\lim_{x \to \infty}\left(\frac{\left(x^{2} - 2 x\right) + 1}{2 x - 1}\right) = \infty
Let's take the limit
so,
horizontal asymptote on the right doesn’t exist
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of (x^2 - 2*x + 1)/(2*x - 1), divided by x at x->+oo and x ->-oo
lim⁡x→−∞((x2−2x)+1x(2x−1))=12\lim_{x \to -\infty}\left(\frac{\left(x^{2} - 2 x\right) + 1}{x \left(2 x - 1\right)}\right) = \frac{1}{2}
Let's take the limit
so,
inclined asymptote equation on the left:
y=x2y = \frac{x}{2}
lim⁡x→∞((x2−2x)+1x(2x−1))=12\lim_{x \to \infty}\left(\frac{\left(x^{2} - 2 x\right) + 1}{x \left(2 x - 1\right)}\right) = \frac{1}{2}
Let's take the limit
so,
inclined asymptote equation on the right:
y=x2y = \frac{x}{2}
Even and odd functions
Let's check, whether the function even or odd by using relations f = f(-x) и f = -f(-x).
So, check:
(x2−2x)+12x−1=x2+2x+1−2x−1\frac{\left(x^{2} - 2 x\right) + 1}{2 x - 1} = \frac{x^{2} + 2 x + 1}{- 2 x - 1}
- No
(x2−2x)+12x−1=−x2+2x+1−2x−1\frac{\left(x^{2} - 2 x\right) + 1}{2 x - 1} = - \frac{x^{2} + 2 x + 1}{- 2 x - 1}
- No
so, the function
not is
neither even, nor odd
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