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Graph for
f(x) = ((2x+3)(x-6))/((x+2)(x-1))

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Answer:

The graph is attached below

Explanation:

The function has three asymptotes. Before we can graph the function, we can find them.

Vertical asymptotes in the values that make the denominator zero.

The denominator becomes zero in:


x = -2\\x = 1\\

Then the vertical asymptotes are the lines


x = -2\\x = 1\\

The horizontal asymptote is found using limits


\lim_(x\to \infty)((2x+3)(x-6))/((x+2)(x-1))

Then:


\lim_(x\to \infty)((2x^2-12x +3x -18))/(x^2-x+2x-2)

We divide the numerator and the denominator between the term of greatest exponent, which in this case is
x ^ 2

The terms of least exponent tend to 0


\lim_(x\to \infty)((2(x^2)/(x^2)-0 +0 -0))/((x^2)/(x^2)-0+0-0)\\\\\lim_(x\to \infty)(2)/(1) = 2\\\\

The function has a horizontal asymptote on y = 2 and has no oblique asymptote

The graph is attached below

Graph for f(x) = ((2x+3)(x-6))/((x+2)(x-1))-example-1
User Kingz
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