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Plotting a function graph/ (x+5)/(x^2-25)

Graphing y = (x+5)/(x^2-25)

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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        x + 5 
f(x) = -------
        2     
       x  - 25
f(x)=x+5x2−25f{\left(x \right)} = \frac{x + 5}{x^{2} - 25} 
f = (x + 5)/(x^2 - 25)
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=−5x_{1} = -5
x2=5x_{2} = 5
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+5x2−25=0\frac{x + 5}{x^{2} - 25} = 0
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to (x + 5)/(x^2 - 25).
5−25+02\frac{5}{-25 + 0^{2}}
The result:
f(0)=−15f{\left(0 \right)} = - \frac{1}{5}
The point:
(0, -1/5)
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+5)(x2−25)2+1x2−25=0- \frac{2 x \left(x + 5\right)}{\left(x^{2} - 25\right)^{2}} + \frac{1}{x^{2} - 25} = 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+5)(4x2x2−25−1))(x2−25)2=0\frac{2 \left(- 2 x + \left(x + 5\right) \left(\frac{4 x^{2}}{x^{2} - 25} - 1\right)\right)}{\left(x^{2} - 25\right)^{2}} = 0
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
x1=−5x_{1} = -5
x2=5x_{2} = 5
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(x+5x2−25)=0\lim_{x \to -\infty}\left(\frac{x + 5}{x^{2} - 25}\right) = 0
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=0y = 0
lim⁡x→∞(x+5x2−25)=0\lim_{x \to \infty}\left(\frac{x + 5}{x^{2} - 25}\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 + 5)/(x^2 - 25), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(x+5x(x2−25))=0\lim_{x \to -\infty}\left(\frac{x + 5}{x \left(x^{2} - 25\right)}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(x+5x(x2−25))=0\lim_{x \to \infty}\left(\frac{x + 5}{x \left(x^{2} - 25\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+5x2−25=5−xx2−25\frac{x + 5}{x^{2} - 25} = \frac{5 - x}{x^{2} - 25}
- No
x+5x2−25=−5−xx2−25\frac{x + 5}{x^{2} - 25} = - \frac{5 - x}{x^{2} - 25}
- No
so, the function
not is
neither even, nor odd
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