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Sagot :
Applying limits, it is found that since [tex]a \neq 0, \frac{a}{2} \neq 0[/tex], and hence the horizontal asymptote is not y = 0, thus not being on the x-axis.
The function is given by:
[tex]F(x) = \frac{(ax - 3)(x - 2)}{2(x + 3)(x - 2)}[/tex]
The horizontal asymptote is the limit of the function as x goes to infinity, hence:
[tex]y = \lim_{x \rigtharrow \infty} F(x) = \lim_{x \rigtharrow \infty} \frac{(ax - 3)(x - 2)}{2(x + 3)(x - 2)}[/tex]
Considering only the terms with the highest exponents:
[tex]y = \lim_{x \rigtharrow \infty} \frac{ax^2}{2x^2} = \frac{a}{2}[/tex]
Since [tex]a \neq 0, \frac{a}{2} \neq 0[/tex], and hence the horizontal asymptote is not y = 0, thus not being on the x-axis.
For more on horizontal asymptotes and limits, you can check https://brainly.com/question/11598999
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