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Sagot :
Step-by-step explanation:
Domain of a rational function is everywhere except where we set vertical asymptotes. or removable discontinues
Here, we have
[tex] \frac{x(2x - 3)}{(x + 2)x} [/tex]
First, notice we have x in both the numerator and denomiator so we have a removable discounties at x.
Since, we don't want x to be 0,
We have a removable discontinuity at x=0
Now, we have
[tex] \frac{2x - 3}{x + 2} [/tex]
We don't want the denomiator be zero because we can't divide by zero.
so
[tex]x + 2 = 0[/tex]
[tex]x = - 2[/tex]
So our domain is
All Real Numbers except-2 and 0.
The vertical asymptors is x=-2.
To find the horinzontal asymptote, notice how the numerator and denomator have the same degree. So this mean we will have a horinzontal asymptoe of
The leading coeffixent of the numerator/ the leading coefficent of the denomiator.
So that becomes
[tex] \frac{2}{1} = 2[/tex]
So we have a horinzontal asymptofe of 2
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