Graphing y = |x+2/x-3|

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       |    2    |
f(x) = |x + - - 3|
       |    x    |

f{\left(x \right)} = \left|{\left(x + \frac{2}{x}\right) - 3}\right|

f = |x + 2/x - 3|

The graph of the function

The domain of the function

The points at which the function is not precisely defined:

x_{1} = 0

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|{\left(x + \frac{2}{x}\right) - 3}\right| = 0

Solve this equation
The points of intersection with the axis X:

Numerical solution

x_{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 - 3|.

\left|{-3 + \frac{2}{0}}\right|

The result:

f{\left(0 \right)} = \infty

sof doesn't intersect Y

Vertical asymptotes

Have:

x_{1} = 0

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|{\left(x + \frac{2}{x}\right) - 3}\right| = \infty

Let's take the limit
so,
horizontal asymptote on the left doesn’t exist

\lim_{x \to \infty} \left|{\left(x + \frac{2}{x}\right) - 3}\right| = \infty

Let's take the limit
so,
horizontal asymptote on the right doesn’t exist

Inclined asymptotes

Inclined asymptote can be found by calculating the limit of |x + 2/x - 3|, divided by x at x->+oo and x ->-oo

\lim_{x \to -\infty}\left(\frac{\left|{\left(x + \frac{2}{x}\right) - 3}\right|}{x}\right) = -1

Let's take the limit
so,
inclined asymptote equation on the left:

y = - x

\lim_{x \to \infty}\left(\frac{\left|{\left(x + \frac{2}{x}\right) - 3}\right|}{x}\right) = 1

Let's take the limit
so,
inclined asymptote equation on the right:

y = x

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|{\left(x + \frac{2}{x}\right) - 3}\right| = \left|{x + 3 + \frac{2}{x}}\right|

- No

\left|{\left(x + \frac{2}{x}\right) - 3}\right| = - \left|{x + 3 + \frac{2}{x}}\right|

- No
so, the function
not is
neither even, nor odd