Question

Graph for exFormula1" title="f(x) = \frac{(2x+3)(x-6)}{(x+2)(x-1)}" alt="f(x) = \frac{(2x+3)(x-6)}{(x+2)(x-1)}" align="absmiddle" class="latex-formula">

Answer 1

Answer:

The graph is attached below

Step-by-step 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

$\lim_{x\to \infty}\frac{(2x+3)(x-6)}{(x+2)(x-1)}$

Then:

$\lim_{x\to \infty}\frac{(2x^2-12x +3x -18)}{x^2-x+2x-2}$

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

$\lim_{x\to \infty}\frac{(2\frac{x^2}{x^2}-0 +0 -0)}{\frac{x^2}{x^2}-0+0-0}\\\\\lim_{x\to \infty}\frac{2}{1} = 2$

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

The graph is attached below