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Sagot :
We have the following:
The first thing is to simplify the expression as much as possible.
In the case of the vertical asymptote: The values that x cannot take is when it makes the denominator zero, therefore we must set the denominator equal to 0 and thus calculate the value of x that will make the function undefined.
2.
[tex]\begin{gathered} \frac{4x^6}{2x-6}=\frac{2\cdot2x^6}{2(x-3)}=\frac{2x^6}{(x-3)} \\ x-3=0 \\ x=3 \end{gathered}[/tex]3.
[tex]\begin{gathered} \frac{6x^2+13x-5}{6x^2-23x+7}=\frac{(3x-1)(2x+5)}{(3x-1)(2x-7)}=\frac{2x+5}{2x-7} \\ 2x-7=0 \\ x=\frac{7}{2} \end{gathered}[/tex]4.
[tex]\begin{gathered} \frac{x+4}{3x^2+11x-4}=\frac{x+4}{(3x-1)(x+4)}=\frac{1}{3x-1} \\ 3x-1=0 \\ x=\frac{1}{3} \end{gathered}[/tex]5.
[tex]\begin{gathered} \frac{-x-4}{x^2-x-20}=\frac{-(x+4)}{(x+4)(x-5)}=\frac{-1}{x-5} \\ x-5=0 \\ x=5 \end{gathered}[/tex]6.
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