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

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

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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       x + 2 + x + 1
f(x) = -------------
           x + 3
f(x)=(x+(x+2))+1x+3f{\left(x \right)} = \frac{\left(x + \left(x + 2\right)\right) + 1}{x + 3} 
f = (x + x + 2 + 1)/(x + 3)
The graph of the function
02468-8-6-4-2-1010-100100
The domain of the function
The points at which the function is not precisely defined:
x1=−3x_{1} = -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+(x+2))+1x+3=0\frac{\left(x + \left(x + 2\right)\right) + 1}{x + 3} = 0
Solve this equation
The points of intersection with the axis X:

Analytical solution
x1=−32x_{1} = - \frac{3}{2}
Numerical solution
x1=−1.5x_{1} = -1.5
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 + 1)/(x + 3).
1+23\frac{1 + 2}{3}
The result:
f(0)=1f{\left(0 \right)} = 1
The point:
(0, 1)
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+3−(x+(x+2))+1(x+3)2=0\frac{2}{x + 3} - \frac{\left(x + \left(x + 2\right)\right) + 1}{\left(x + 3\right)^{2}} = 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(−2+2x+3x+3)(x+3)2=0\frac{2 \left(-2 + \frac{2 x + 3}{x + 3}\right)}{\left(x + 3\right)^{2}} = 0
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
x1=−3x_{1} = -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+(x+2))+1x+3)=2\lim_{x \to -\infty}\left(\frac{\left(x + \left(x + 2\right)\right) + 1}{x + 3}\right) = 2
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=2y = 2
lim⁡x→∞((x+(x+2))+1x+3)=2\lim_{x \to \infty}\left(\frac{\left(x + \left(x + 2\right)\right) + 1}{x + 3}\right) = 2
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=2y = 2
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of (x + 2 + x + 1)/(x + 3), divided by x at x->+oo and x ->-oo
lim⁡x→−∞((x+(x+2))+1x(x+3))=0\lim_{x \to -\infty}\left(\frac{\left(x + \left(x + 2\right)\right) + 1}{x \left(x + 3\right)}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞((x+(x+2))+1x(x+3))=0\lim_{x \to \infty}\left(\frac{\left(x + \left(x + 2\right)\right) + 1}{x \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))+1x+3=3−2x3−x\frac{\left(x + \left(x + 2\right)\right) + 1}{x + 3} = \frac{3 - 2 x}{3 - x}
- No
(x+(x+2))+1x+3=−3−2x3−x\frac{\left(x + \left(x + 2\right)\right) + 1}{x + 3} = - \frac{3 - 2 x}{3 - x}
- No
so, the function
not is
neither even, nor odd
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