Rational Functions Warmup Practice Problems Online

What is the y-intercept of the graph of \[y=\dfrac{(x+3)(x-8)}{(x-6)(x+2)}\]

\[\dfrac{(2x+2)}{(x+2)(x+7)} = 1\] What are the possible values of \(x\) ?

\[x=-3\] \(x=-3\) or \(x=2\) \(x=-3\) or \(x=-4\) \[x=-4\] \(x=2\) or \(x=-4\) \[x=2\] There are no values of \(x\) which satisfy this expression

What is the range of the function defined by \[f(x) = \dfrac{(x+2)(x+2)(x+2)}{(x+2)} ?\]

\[\mathbb{R}\] \[[2,\infty)\] \[[-2,\infty)\] \[(0,\infty)\] \[[0,\infty)\] \[(-\infty,0]\] \[(-\infty,0)\]

The red line above is a graph of \(y=f(x).\) The graph travels through the points \((3,0)\) and \((0, -2.25).\)

The blue line shows the horizontal asymptote of \(f(x)\).

What is a possible definition for \(f(x)\)?

\[f(x) = \dfrac{(x-3)}{(x+2)(x-2)(x-2)}\] \[f(x) = \dfrac{(x-3)}{(x-2)(x-2)}\] \[f(x) = \dfrac{(x-3)}{2(x+2)(x-2)}\] \[f(x) = \dfrac{(x-3)(x-3)}{(x+2)(x-2)}\] \[f(x) = \dfrac{(x-3)(x-2)}{(x+2)(x-2)}\] \[f(x) = \dfrac{(x-3)(x+3)}{(x+2)(x-2)}\] \[f(x) = \dfrac{(x-3)}{(x+2)(x-2)}\]

Where are the vertical asymptotes of the function \(f(x) = \dfrac{(x-6)(x-4)}{2(x-4)(x+3)}\) ?

\[x=-3\] \[x=-4\] \(x=4\) and \(x=6\) \(x=-3\) and \(x=4\) \(x=-6\) and \(x=-4\) This function has no vertical asymptotes \(x=3\) and \(x=-4\)

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