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Plotting a function graph/ (1-x)/((x+2)*(x-3))

Graphing y = (1-x)/((x+2)*(x-3))

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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            1 - x     
f(x) = ---------------
       (x + 2)*(x - 3)
f(x)=1−x(x−3)(x+2)f{\left(x \right)} = \frac{1 - x}{\left(x - 3\right) \left(x + 2\right)} 
f = (1 - x)/(((x - 3)*(x + 2)))
The graph of the function
02468-8-6-4-2-1010-2525
The domain of the function
The points at which the function is not precisely defined:
x1=−2x_{1} = -2
x2=3x_{2} = 3
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:
1−x(x−3)(x+2)=0\frac{1 - x}{\left(x - 3\right) \left(x + 2\right)} = 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 (1 - x)/(((x + 2)*(x - 3))).
1−0(−3)2\frac{1 - 0}{\left(-3\right) 2}
The result:
f(0)=−16f{\left(0 \right)} = - \frac{1}{6}
The point:
(0, -1/6)
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
(1−2x)(1−x)(x−3)2(x+2)2−1(x−3)(x+2)=0\frac{\left(1 - 2 x\right) \left(1 - x\right)}{\left(x - 3\right)^{2} \left(x + 2\right)^{2}} - \frac{1}{\left(x - 3\right) \left(x + 2\right)} = 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
4x−(x−1)((2x−1)(1x+2+1x−3)−2+2x−1x+2+2x−1x−3)−2(x−3)2(x+2)2=0\frac{4 x - \left(x - 1\right) \left(\left(2 x - 1\right) \left(\frac{1}{x + 2} + \frac{1}{x - 3}\right) - 2 + \frac{2 x - 1}{x + 2} + \frac{2 x - 1}{x - 3}\right) - 2}{\left(x - 3\right)^{2} \left(x + 2\right)^{2}} = 0
Solve this equation
The roots of this equation
x1=−183+1+123x_{1} = - \sqrt[3]{18} + 1 + \sqrt[3]{12}
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
x1=−2x_{1} = -2
x2=3x_{2} = 3

lim⁡x→−2−(4x−(x−1)((2x−1)(1x+2+1x−3)−2+2x−1x+2+2x−1x−3)−2(x−3)2(x+2)2)=∞\lim_{x \to -2^-}\left(\frac{4 x - \left(x - 1\right) \left(\left(2 x - 1\right) \left(\frac{1}{x + 2} + \frac{1}{x - 3}\right) - 2 + \frac{2 x - 1}{x + 2} + \frac{2 x - 1}{x - 3}\right) - 2}{\left(x - 3\right)^{2} \left(x + 2\right)^{2}}\right) = \infty
lim⁡x→−2+(4x−(x−1)((2x−1)(1x+2+1x−3)−2+2x−1x+2+2x−1x−3)−2(x−3)2(x+2)2)=−∞\lim_{x \to -2^+}\left(\frac{4 x - \left(x - 1\right) \left(\left(2 x - 1\right) \left(\frac{1}{x + 2} + \frac{1}{x - 3}\right) - 2 + \frac{2 x - 1}{x + 2} + \frac{2 x - 1}{x - 3}\right) - 2}{\left(x - 3\right)^{2} \left(x + 2\right)^{2}}\right) = -\infty
- the limits are not equal, so
x1=−2x_{1} = -2
- is an inflection point
lim⁡x→3−(4x−(x−1)((2x−1)(1x+2+1x−3)−2+2x−1x+2+2x−1x−3)−2(x−3)2(x+2)2)=∞\lim_{x \to 3^-}\left(\frac{4 x - \left(x - 1\right) \left(\left(2 x - 1\right) \left(\frac{1}{x + 2} + \frac{1}{x - 3}\right) - 2 + \frac{2 x - 1}{x + 2} + \frac{2 x - 1}{x - 3}\right) - 2}{\left(x - 3\right)^{2} \left(x + 2\right)^{2}}\right) = \infty
lim⁡x→3+(4x−(x−1)((2x−1)(1x+2+1x−3)−2+2x−1x+2+2x−1x−3)−2(x−3)2(x+2)2)=−∞\lim_{x \to 3^+}\left(\frac{4 x - \left(x - 1\right) \left(\left(2 x - 1\right) \left(\frac{1}{x + 2} + \frac{1}{x - 3}\right) - 2 + \frac{2 x - 1}{x + 2} + \frac{2 x - 1}{x - 3}\right) - 2}{\left(x - 3\right)^{2} \left(x + 2\right)^{2}}\right) = -\infty
- the limits are not equal, so
x2=3x_{2} = 3
- is an inflection point

Сonvexity and concavity intervals:
Let’s find the intervals where the function is convex or concave, for this look at the behaviour of the function at the inflection points:
Concave at the intervals
[−183+1+123,∞)\left[- \sqrt[3]{18} + 1 + \sqrt[3]{12}, \infty\right)
Convex at the intervals
(−∞,−183+1+123]\left(-\infty, - \sqrt[3]{18} + 1 + \sqrt[3]{12}\right]
Vertical asymptotes
Have:
x1=−2x_{1} = -2
x2=3x_{2} = 3
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(1−x(x−3)(x+2))=0\lim_{x \to -\infty}\left(\frac{1 - x}{\left(x - 3\right) \left(x + 2\right)}\right) = 0
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=0y = 0
lim⁡x→∞(1−x(x−3)(x+2))=0\lim_{x \to \infty}\left(\frac{1 - x}{\left(x - 3\right) \left(x + 2\right)}\right) = 0
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=0y = 0
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of (1 - x)/(((x + 2)*(x - 3))), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(1(x−3)(x+2)(1−x)x)=0\lim_{x \to -\infty}\left(\frac{\frac{1}{\left(x - 3\right) \left(x + 2\right)} \left(1 - x\right)}{x}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(1(x−3)(x+2)(1−x)x)=0\lim_{x \to \infty}\left(\frac{\frac{1}{\left(x - 3\right) \left(x + 2\right)} \left(1 - x\right)}{x}\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:
1−x(x−3)(x+2)=x+1(2−x)(−x−3)\frac{1 - x}{\left(x - 3\right) \left(x + 2\right)} = \frac{x + 1}{\left(2 - x\right) \left(- x - 3\right)}
- No
1−x(x−3)(x+2)=−x+1(2−x)(−x−3)\frac{1 - x}{\left(x - 3\right) \left(x + 2\right)} = - \frac{x + 1}{\left(2 - x\right) \left(- x - 3\right)}
- No
so, the function
not is
neither even, nor odd
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