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Plotting a function graph/ (x+2)/(x^2-9)

Graphing y = (x+2)/(x^2-9)

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
       x + 2 
f(x) = ------
        2    
       x  - 9
f(x)=x+2x2−9f{\left(x \right)} = \frac{x + 2}{x^{2} - 9} 
f = (x + 2)/(x^2 - 9)
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=−3x_{1} = -3
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:
x+2x2−9=0\frac{x + 2}{x^{2} - 9} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=−2x_{1} = -2
Numerical solution
x1=−2x_{1} = -2
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^2 - 9).
2−9+02\frac{2}{-9 + 0^{2}}
The result:
f(0)=−29f{\left(0 \right)} = - \frac{2}{9}
The point:
(0, -2/9)
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(x+2)(x2−9)2+1x2−9=0- \frac{2 x \left(x + 2\right)}{\left(x^{2} - 9\right)^{2}} + \frac{1}{x^{2} - 9} = 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(−2x+(x+2)(4x2x2−9−1))(x2−9)2=0\frac{2 \left(- 2 x + \left(x + 2\right) \left(\frac{4 x^{2}}{x^{2} - 9} - 1\right)\right)}{\left(x^{2} - 9\right)^{2}} = 0
Solve this equation
The roots of this equation
x1=−2−53+523x_{1} = -2 - \sqrt[3]{5} + 5^{\frac{2}{3}}
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=3x_{2} = 3

lim⁡x→−3−(2(−2x+(x+2)(4x2x2−9−1))(x2−9)2)=−∞\lim_{x \to -3^-}\left(\frac{2 \left(- 2 x + \left(x + 2\right) \left(\frac{4 x^{2}}{x^{2} - 9} - 1\right)\right)}{\left(x^{2} - 9\right)^{2}}\right) = -\infty
lim⁡x→−3+(2(−2x+(x+2)(4x2x2−9−1))(x2−9)2)=∞\lim_{x \to -3^+}\left(\frac{2 \left(- 2 x + \left(x + 2\right) \left(\frac{4 x^{2}}{x^{2} - 9} - 1\right)\right)}{\left(x^{2} - 9\right)^{2}}\right) = \infty
- the limits are not equal, so
x1=−3x_{1} = -3
- is an inflection point
lim⁡x→3−(2(−2x+(x+2)(4x2x2−9−1))(x2−9)2)=−∞\lim_{x \to 3^-}\left(\frac{2 \left(- 2 x + \left(x + 2\right) \left(\frac{4 x^{2}}{x^{2} - 9} - 1\right)\right)}{\left(x^{2} - 9\right)^{2}}\right) = -\infty
lim⁡x→3+(2(−2x+(x+2)(4x2x2−9−1))(x2−9)2)=∞\lim_{x \to 3^+}\left(\frac{2 \left(- 2 x + \left(x + 2\right) \left(\frac{4 x^{2}}{x^{2} - 9} - 1\right)\right)}{\left(x^{2} - 9\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
(−∞,−2−53+523]\left(-\infty, -2 - \sqrt[3]{5} + 5^{\frac{2}{3}}\right]
Convex at the intervals
[−2−53+523,∞)\left[-2 - \sqrt[3]{5} + 5^{\frac{2}{3}}, \infty\right)
Vertical asymptotes
Have:
x1=−3x_{1} = -3
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→−∞(x+2x2−9)=0\lim_{x \to -\infty}\left(\frac{x + 2}{x^{2} - 9}\right) = 0
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=0y = 0
lim⁡x→∞(x+2x2−9)=0\lim_{x \to \infty}\left(\frac{x + 2}{x^{2} - 9}\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 + 2)/(x^2 - 9), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(x+2x(x2−9))=0\lim_{x \to -\infty}\left(\frac{x + 2}{x \left(x^{2} - 9\right)}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(x+2x(x2−9))=0\lim_{x \to \infty}\left(\frac{x + 2}{x \left(x^{2} - 9\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+2x2−9=2−xx2−9\frac{x + 2}{x^{2} - 9} = \frac{2 - x}{x^{2} - 9}
- No
x+2x2−9=−2−xx2−9\frac{x + 2}{x^{2} - 9} = - \frac{2 - x}{x^{2} - 9}
- No
so, the function
not is
neither even, nor odd
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