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

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

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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       2*x - 3
f(x) = -------
        x + 1
f(x)=2x−3x+1f{\left(x \right)} = \frac{2 x - 3}{x + 1} 
f = (2*x - 3)/(x + 1)
The graph of the function
02468-8-6-4-2-1010-200200
The domain of the function
The points at which the function is not precisely defined:
x1=−1x_{1} = -1
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:
2x−3x+1=0\frac{2 x - 3}{x + 1} = 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 (2*x - 3)/(x + 1).
−3+0⋅21\frac{-3 + 0 \cdot 2}{1}
The result:
f(0)=−3f{\left(0 \right)} = -3
The point:
(0, -3)
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+1−2x−3(x+1)2=0\frac{2}{x + 1} - \frac{2 x - 3}{\left(x + 1\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+1)(x+1)2=0\frac{2 \left(-2 + \frac{2 x - 3}{x + 1}\right)}{\left(x + 1\right)^{2}} = 0
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
x1=−1x_{1} = -1
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
lim⁡x→−∞(2x−3x+1)=2\lim_{x \to -\infty}\left(\frac{2 x - 3}{x + 1}\right) = 2
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=2y = 2
lim⁡x→∞(2x−3x+1)=2\lim_{x \to \infty}\left(\frac{2 x - 3}{x + 1}\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 (2*x - 3)/(x + 1), divided by x at x->+oo and x ->-oo
lim⁡x→−∞(2x−3x(x+1))=0\lim_{x \to -\infty}\left(\frac{2 x - 3}{x \left(x + 1\right)}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡x→∞(2x−3x(x+1))=0\lim_{x \to \infty}\left(\frac{2 x - 3}{x \left(x + 1\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:
2x−3x+1=−2x−31−x\frac{2 x - 3}{x + 1} = \frac{- 2 x - 3}{1 - x}
- No
2x−3x+1=−−2x−31−x\frac{2 x - 3}{x + 1} = - \frac{- 2 x - 3}{1 - x}
- No
so, the function
not is
neither even, nor odd
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