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Graphing y = abs((2x-3)/(x+2))

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The graph:

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Intersection points:

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Piecewise:

The solution

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       |2*x - 3|
f(x) = |-------|
       | x + 2 |
$$f{\left(x \right)} = \left|{\frac{2 x - 3}{x + 2}}\right|$$
f = Abs((2*x - 3)/(x + 2))
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = -2$$
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:
$$\left|{\frac{2 x - 3}{x + 2}}\right| = 0$$
Solve this equation
The points of intersection with the axis X:

Numerical solution
$$x_{1} = 1.5$$
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to Abs((2*x - 3)/(x + 2)).
$$\left|{\frac{-3 + 0 \cdot 2}{2}}\right|$$
The result:
$$f{\left(0 \right)} = \frac{3}{2}$$
The point:
(0, 3/2)
Extrema of the function
In order to find the extrema, we need to solve the equation
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
$$\frac{d}{d x} f{\left(x \right)} = $$
the first derivative
$$\left(\frac{2}{x + 2} - \frac{2 x - 3}{\left(x + 2\right)^{2}}\right) \operatorname{sign}{\left(\frac{2 x - 3}{x + 2} \right)} = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
Vertical asymptotes
Have:
$$x_{1} = -2$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty} \left|{\frac{2 x - 3}{x + 2}}\right| = 2$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 2$$
$$\lim_{x \to \infty} \left|{\frac{2 x - 3}{x + 2}}\right| = 2$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 2$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of Abs((2*x - 3)/(x + 2)), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\left|{\frac{2 x - 3}{x + 2}}\right|}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\left|{\frac{2 x - 3}{x + 2}}\right|}{x}\right) = 0$$
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:
$$\left|{\frac{2 x - 3}{x + 2}}\right| = \left|{\frac{2 x + 3}{x - 2}}\right|$$
- No
$$\left|{\frac{2 x - 3}{x + 2}}\right| = - \left|{\frac{2 x + 3}{x - 2}}\right|$$
- No
so, the function
not is
neither even, nor odd