Graph for [tex]f(x) = \frac{(2x+3)(x-6)}{(x+2)(x-1)}[/tex]

Answer:The graph is attached belowStep-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:[tex]x = -2\\x = 1\\[/tex]Then the vertical asymptotes are the lines[tex]x = -2\\x = 1\\[/tex]The horizontal asymptote is found using limits[tex]\lim_{x\to \infty}\frac{(2x+3)(x-6)}{(x+2)(x-1)}[/tex]Then:[tex]\lim_{x\to \infty}\frac{(2x^2-12x +3x -18)}{x^2-x+2x-2}[/tex]We divide the numerator and the denominator between the term of greatest exponent, which in this case is [tex]x ^ 2[/tex]The terms of least exponent tend to 0[tex]\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\\\\[/tex]The function has a horizontal asymptote on y = 2 and has no oblique asymptoteThe graph is attached below