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

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

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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       x - 2*x + 1
f(x) = -----------
          x - 2
f(x)=(−2x+x)+1x−2f{\left(x \right)} = \frac{\left(- 2 x + x\right) + 1}{x - 2} 
f = (-2*x + x + 1)/(x - 2)
The graph of the function
02468-8-6-4-2-1010-5050
The domain of the function
The points at which the function is not precisely defined:
x1=2x_{1} = 2
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:
(−2x+x)+1x−2=0\frac{\left(- 2 x + x\right) + 1}{x - 2} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=1x_{1} = 1
Numerical solution
x1=1x_{1} = 1
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to (x - 2*x + 1)/(x - 2).
1−0−2\frac{1 - 0}{-2}
The result:
f(0)=−12f{\left(0 \right)} = - \frac{1}{2}
The point:
(0, -1/2)
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
−1x−2−(−2x+x)+1(x−2)2=0- \frac{1}{x - 2} - \frac{\left(- 2 x + x\right) + 1}{\left(x - 2\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−xx−2+1)(x−2)2=0\frac{2 \left(\frac{1 - x}{x - 2} + 1\right)}{\left(x - 2\right)^{2}} = 0
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
x1=2x_{1} = 2
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞((−2x+x)+1x−2)=−1\lim_{x \to -\infty}\left(\frac{\left(- 2 x + x\right) + 1}{x - 2}\right) = -1
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=−1y = -1
lim⁡x→∞((−2x+x)+1x−2)=−1\lim_{x \to \infty}\left(\frac{\left(- 2 x + x\right) + 1}{x - 2}\right) = -1
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=−1y = -1
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of (x - 2*x + 1)/(x - 2), divided by x at x->+oo and x ->-oo
lim⁡x→−∞((−2x+x)+1x(x−2))=0\lim_{x \to -\infty}\left(\frac{\left(- 2 x + x\right) + 1}{x \left(x - 2\right)}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞((−2x+x)+1x(x−2))=0\lim_{x \to \infty}\left(\frac{\left(- 2 x + x\right) + 1}{x \left(x - 2\right)}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
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:
(−2x+x)+1x−2=x+1−x−2\frac{\left(- 2 x + x\right) + 1}{x - 2} = \frac{x + 1}{- x - 2}
- No
(−2x+x)+1x−2=−x+1−x−2\frac{\left(- 2 x + x\right) + 1}{x - 2} = - \frac{x + 1}{- x - 2}
- No
so, the function
not is
neither even, nor odd
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