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

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

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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        2          
       x  - 2*x + 1
f(x) = ------------
          x - 1
f(x)=(x2−2x)+1x−1f{\left(x \right)} = \frac{\left(x^{2} - 2 x\right) + 1}{x - 1} 
f = (x^2 - 2*x + 1)/(x - 1)
The graph of the function
02468-8-6-4-2-1010-2020
The domain of the function
The points at which the function is not precisely defined:
x1=1x_{1} = 1
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)+1x−1=0\frac{\left(x^{2} - 2 x\right) + 1}{x - 1} = 0
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
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)/(x - 1).
(02−0)+1−1\frac{\left(0^{2} - 0\right) + 1}{-1}
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−2x−1−(x2−2x)+1(x−1)2=0\frac{2 x - 2}{x - 1} - \frac{\left(x^{2} - 2 x\right) + 1}{\left(x - 1\right)^{2}} = 0
Solve this equation
Solutions are not found,
function may have no extrema
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(−1+x2−2x+1(x−1)2)x−1=0\frac{2 \left(-1 + \frac{x^{2} - 2 x + 1}{\left(x - 1\right)^{2}}\right)}{x - 1} = 0
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
x1=1x_{1} = 1
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)+1x−1)=−∞\lim_{x \to -\infty}\left(\frac{\left(x^{2} - 2 x\right) + 1}{x - 1}\right) = -\infty
Let's take the limit
so,
horizontal asymptote on the left doesn’t exist
lim⁡x→∞((x2−2x)+1x−1)=∞\lim_{x \to \infty}\left(\frac{\left(x^{2} - 2 x\right) + 1}{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)/(x - 1), divided by x at x->+oo and x ->-oo
lim⁡x→−∞((x2−2x)+1x(x−1))=1\lim_{x \to -\infty}\left(\frac{\left(x^{2} - 2 x\right) + 1}{x \left(x - 1\right)}\right) = 1
Let's take the limit
so,
inclined asymptote equation on the left:
y=xy = x
lim⁡x→∞((x2−2x)+1x(x−1))=1\lim_{x \to \infty}\left(\frac{\left(x^{2} - 2 x\right) + 1}{x \left(x - 1\right)}\right) = 1
Let's take the limit
so,
inclined asymptote equation on the right:
y=xy = x
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)+1x−1=x2+2x+1−x−1\frac{\left(x^{2} - 2 x\right) + 1}{x - 1} = \frac{x^{2} + 2 x + 1}{- x - 1}
- No
(x2−2x)+1x−1=−x2+2x+1−x−1\frac{\left(x^{2} - 2 x\right) + 1}{x - 1} = - \frac{x^{2} + 2 x + 1}{- x - 1}
- No
so, the function
not is
neither even, nor odd
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