graph the rational function.\n g(x)=\frac{-3x + 12}{x^{2}-5x + 4} \nstart by drawing the asymptotes (if…

Explanation:

Step1: Factor the function

First, factor the numerator and denominator. The numerator (-3x + 12=-3(x - 4)). The denominator (x^{2}-5x + 4=(x - 1)(x - 4)). So (g(x)=\frac{-3(x - 4)}{(x - 1)(x - 4)}), and after canceling out the non - zero factor ((x - 4)) (for (x\neq4)), we get (g(x)=\frac{-3}{x - 1},x\neq4).

Step2: Find the vertical asymptote

Set the denominator of the simplified function equal to zero. For (y=\frac{-3}{x - 1}), when (x-1 = 0), i.e., (x = 1), there is a vertical asymptote.

Step3: Find the horizontal asymptote

Since the degree of the numerator is 0 and the degree of the denominator is 1 ((n=0,m = 1) where (n) is the degree of the numerator and (m) is the degree of the denominator), the horizontal asymptote is (y = 0).

Step4: Find the hole

Since we canceled out the factor ((x - 4)) in the original function, there is a hole at (x = 4). To find the (y) - value of the hole, substitute (x = 4) into the simplified function (y=\frac{-3}{x - 1}), we get (y=-1). So the hole is at the point ((4,-1)).

Step5: Plot points

For (x=0), (y=\frac{-3}{0 - 1}=3). For (x = 2), (y=\frac{-3}{2 - 1}=-3). For (x=5), (y=\frac{-3}{5 - 1}=-\frac{3}{4}). For (x=-1), (y=\frac{-3}{-1 - 1}=\frac{3}{2}).

To graph: Draw a vertical asymptote (x = 1), a horizontal asymptote (y = 0), plot the hole at the point ((4,-1)) as a hollow dot, and plot the points ((0,3),(2,-3),(5,-\frac{3}{4}),(-1,\frac{3}{2})) and then draw the two branches of the rational - function graph approaching the asymptotes.

Answer:

Graph with vertical asymptote (x = 1), horizontal asymptote (y = 0), hole at ((4,-1)) and points ((0,3),(2,-3),(5,-\frac{3}{4}),(-1,\frac{3}{2})) plotted.