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

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

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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       -2*(x + 1)
f(x) = ----------
         2       
        x  + 2*x
f(x)=−2(x+1)x2+2xf{\left(x \right)} = - \frac{2 \left(x + 1\right)}{x^{2} + 2 x} 
f = -2*(x + 1)/(x^2 + 2*x)
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
x2=0x_{2} = 0
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:
−2(x+1)x2+2x=0- \frac{2 \left(x + 1\right)}{x^{2} + 2 x} = 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 -2*(x + 1)/(x^2 + 2*x).
−2⋅(0+1)02+2⋅0- \frac{2 \cdot \left(0 + 1\right)}{0^{2} + 2 \cdot 0}
The result:
f(0)=∞~f{\left(0 \right)} = \tilde{\infty}
sof doesn't intersect Y
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
−2(−2x−2)(x+1)(x2+2x)2−2x2+2x=0- \frac{2 \left(- 2 x - 2\right) \left(x + 1\right)}{\left(x^{2} + 2 x\right)^{2}} - \frac{2}{x^{2} + 2 x} = 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
4⋅(3−4(x+1)2x(x+2))(x+1)x2(x+2)2=0\frac{4 \cdot \left(3 - \frac{4 \left(x + 1\right)^{2}}{x \left(x + 2\right)}\right) \left(x + 1\right)}{x^{2} \left(x + 2\right)^{2}} = 0
Solve this equation
The roots of this equation
x1=−1x_{1} = -1
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=0x_{2} = 0

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