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

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

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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              x       
f(x) = ---------------
       (x - 2)*(x + 3)
f(x)=x(x−2)(x+3)f{\left(x \right)} = \frac{x}{\left(x - 2\right) \left(x + 3\right)} 
f = x/(((x - 2)*(x + 3)))
The graph of the function
0.05.00.51.01.52.02.53.03.54.04.5-50000100000
The domain of the function
The points at which the function is not precisely defined:
x1=−3x_{1} = -3
x2=2x_{2} = 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:
x(x−2)(x+3)=0\frac{x}{\left(x - 2\right) \left(x + 3\right)} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=0x_{1} = 0
Numerical solution
x1=0x_{1} = 0
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to x/(((x - 2)*(x + 3))).
0(−1)2⋅3\frac{0}{\left(-1\right) 2 \cdot 3}
The result:
f(0)=0f{\left(0 \right)} = 0
The point:
(0, 0)
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
x(−2x−1)(x−2)2(x+3)2+1(x−2)(x+3)=0\frac{x \left(- 2 x - 1\right)}{\left(x - 2\right)^{2} \left(x + 3\right)^{2}} + \frac{1}{\left(x - 2\right) \left(x + 3\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
x((2x+1)(1x+3+1x−2)−2+2x+1x+3+2x+1x−2)−4x−2(x−2)2(x+3)2=0\frac{x \left(\left(2 x + 1\right) \left(\frac{1}{x + 3} + \frac{1}{x - 2}\right) - 2 + \frac{2 x + 1}{x + 3} + \frac{2 x + 1}{x - 2}\right) - 4 x - 2}{\left(x - 2\right)^{2} \left(x + 3\right)^{2}} = 0
Solve this equation
The roots of this equation
x1=−183+123x_{1} = - \sqrt[3]{18} + \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=−3x_{1} = -3
x2=2x_{2} = 2

lim⁡x→−3−(x((2x+1)(1x+3+1x−2)−2+2x+1x+3+2x+1x−2)−4x−2(x−2)2(x+3)2)=−∞\lim_{x \to -3^-}\left(\frac{x \left(\left(2 x + 1\right) \left(\frac{1}{x + 3} + \frac{1}{x - 2}\right) - 2 + \frac{2 x + 1}{x + 3} + \frac{2 x + 1}{x - 2}\right) - 4 x - 2}{\left(x - 2\right)^{2} \left(x + 3\right)^{2}}\right) = -\infty
lim⁡x→−3+(x((2x+1)(1x+3+1x−2)−2+2x+1x+3+2x+1x−2)−4x−2(x−2)2(x+3)2)=∞\lim_{x \to -3^+}\left(\frac{x \left(\left(2 x + 1\right) \left(\frac{1}{x + 3} + \frac{1}{x - 2}\right) - 2 + \frac{2 x + 1}{x + 3} + \frac{2 x + 1}{x - 2}\right) - 4 x - 2}{\left(x - 2\right)^{2} \left(x + 3\right)^{2}}\right) = \infty
- the limits are not equal, so
x1=−3x_{1} = -3
- is an inflection point
lim⁡x→2−(x((2x+1)(1x+3+1x−2)−2+2x+1x+3+2x+1x−2)−4x−2(x−2)2(x+3)2)=−∞\lim_{x \to 2^-}\left(\frac{x \left(\left(2 x + 1\right) \left(\frac{1}{x + 3} + \frac{1}{x - 2}\right) - 2 + \frac{2 x + 1}{x + 3} + \frac{2 x + 1}{x - 2}\right) - 4 x - 2}{\left(x - 2\right)^{2} \left(x + 3\right)^{2}}\right) = -\infty
lim⁡x→2+(x((2x+1)(1x+3+1x−2)−2+2x+1x+3+2x+1x−2)−4x−2(x−2)2(x+3)2)=∞\lim_{x \to 2^+}\left(\frac{x \left(\left(2 x + 1\right) \left(\frac{1}{x + 3} + \frac{1}{x - 2}\right) - 2 + \frac{2 x + 1}{x + 3} + \frac{2 x + 1}{x - 2}\right) - 4 x - 2}{\left(x - 2\right)^{2} \left(x + 3\right)^{2}}\right) = \infty
- the limits are not equal, so
x2=2x_{2} = 2
- 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+123]\left(-\infty, - \sqrt[3]{18} + \sqrt[3]{12}\right]
Convex at the intervals
[−183+123,∞)\left[- \sqrt[3]{18} + \sqrt[3]{12}, \infty\right)
Vertical asymptotes
Have:
x1=−3x_{1} = -3
x2=2x_{2} = 2
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(x(x−2)(x+3))=0\lim_{x \to -\infty}\left(\frac{x}{\left(x - 2\right) \left(x + 3\right)}\right) = 0
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=0y = 0
lim⁡x→∞(x(x−2)(x+3))=0\lim_{x \to \infty}\left(\frac{x}{\left(x - 2\right) \left(x + 3\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 x/(((x - 2)*(x + 3))), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(1(x−2)(x+3))=0\lim_{x \to -\infty}\left(\frac{1}{\left(x - 2\right) \left(x + 3\right)}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(1(x−2)(x+3))=0\lim_{x \to \infty}\left(\frac{1}{\left(x - 2\right) \left(x + 3\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:
x(x−2)(x+3)=−x(3−x)(−x−2)\frac{x}{\left(x - 2\right) \left(x + 3\right)} = - \frac{x}{\left(3 - x\right) \left(- x - 2\right)}
- No
x(x−2)(x+3)=x(3−x)(−x−2)\frac{x}{\left(x - 2\right) \left(x + 3\right)} = \frac{x}{\left(3 - x\right) \left(- x - 2\right)}
- No
so, the function
not is
neither even, nor odd
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