The function [tex]f(x)=\dfrac{\left(-2x+4\right)}{x-6}[/tex] will have a domain of [tex](-\infty,6)\cup(6,+\infty)[/tex], the function [tex]g(x)=\dfrac{\left(x^{2}-2x\right)}{x^{2}+x-6}[/tex] will have a domain of [tex](-\infty,-3)\cup(-3,2)\cup(2,+\infty)[/tex], and the function [tex]h(x)=\dfrac{\left(x^{2}-2x\right)}{x-6}[/tex] will have a domain of [tex](-\infty,6)\cup(6,+\infty)[/tex].
Given information:
The given functions are:
[tex]f(x)=\dfrac{\left(-2x+4\right)}{x-6}[/tex]
[tex]g(x)=\dfrac{\left(x^{2}-2x\right)}{x^{2}+x-6}[/tex]
[tex]h(x)=\dfrac{\left(x^{2}-2x\right)}{x-6}[/tex]
For a rational function, the denominator should not be equal to zero.
So, the domain of the first function will be,
[tex]x-6\neq0\\x\neq6\\(-\infty,6)\cup(6,+\infty)[/tex]
The function [tex]f(x)=\dfrac{\left(-2x+4\right)}{x-6}[/tex] will form a vertical asymptote at [tex]x=6[/tex].
The domain of the second function will be,
[tex]x^2+x-6\neq0\\x^2+3x-2x-6\neq0\\(x+3)(x-2)\neq0\\x\neq2, -3\\(-\infty,-3)\cup(-3,2)\cup(2,+\infty)[/tex]
The function [tex]g(x)=\dfrac{\left(x^{2}-2x\right)}{x^{2}+x-6}[/tex] will form a vertical asymptote at [tex]x=-3[/tex] and a horizontal asymptote at [tex]y=1[/tex].
The domain of third function h(x) will be,
[tex]x-6\neq0\\x\neq6\\(-\infty,6)\cup(6,+\infty)[/tex]
Now, the third function [tex]h(x)=\dfrac{\left(x^{2}-2x\right)}{x-6}[/tex] will form an oblique asymptote, and a vertical asymptote at [tex]x=6[/tex].
The graph of each function is attached to the image.
Therefore, the function [tex]f(x)=\dfrac{\left(-2x+4\right)}{x-6}[/tex] will have a domain of [tex](-\infty,6)\cup(6,+\infty)[/tex], the function [tex]g(x)=\dfrac{\left(x^{2}-2x\right)}{x^{2}+x-6}[/tex] will have a domain of [tex](-\infty,-3)\cup(-3,2)\cup(2,+\infty)[/tex], and the function [tex]h(x)=\dfrac{\left(x^{2}-2x\right)}{x-6}[/tex] will have a domain of [tex](-\infty,6)\cup(6,+\infty)[/tex].
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https://brainly.com/question/15099991
