Questions: Graph the rational function. f(x)=(2x^2-6x+3)/(3x-9) Start by drawing the asymptotes. Then plot two points on each piece of the graph. Finally, click on the graph-a-function button.

Solution Steps

We can simplify the given function $f(x) = \frac{2x^2 - 6x + 3}{3x - 9}$ by factoring out a 3 in the denominator: $f(x) = \frac{2x^2 - 6x + 3}{3(x - 3)}$

A vertical asymptote occurs when the denominator is zero and the numerator is non-zero. Setting the denominator equal to zero, we get: $3(x - 3) = 0$ $x - 3 = 0$ $x = 3$ Since the numerator is not zero at $x=3$, there is a vertical asymptote at $x = 3$.

Since the degree of the numerator (2) is greater than the degree of the denominator (1), there is a slant (oblique) asymptote. To find the slant asymptote, we perform polynomial long division. Dividing $2x^2 - 6x + 3$ by $3x - 9$, we get $\frac{2}{3}x - \frac{2}{3}$ with a remainder of -3. So, $f(x) = \frac{2}{3}x - \frac{2}{3} + \frac{-3}{3x-9} = \frac{2}{3}x - \frac{2}{3} - \frac{1}{x-3}$. As x approaches infinity, the term $\frac{-1}{x-3}$ approaches 0, thus the slant asymptote is $y = \frac{2}{3}x - \frac{2}{3}$.

We need to find two points on either side of the vertical asymptote $x = 3$.

For $x < 3$: Let $x=0$. $f(0) = \frac{3}{-9} = -\frac{1}{3}$. So we have the point $(0, -\frac{1}{3})$. Let $x=2$. $f(2) = \frac{2(4)-6(2)+3}{3(2)-9} = \frac{-1}{-3} = \frac{1}{3}$. So we have the point $(2, \frac{1}{3})$.

For $x > 3$: Let $x=4$. $f(4) = \frac{2(16)-6(4)+3}{3(4)-9} = \frac{11}{3}$. So we have the point $(4, \frac{11}{3})$. Let $x=5$. $f(5) = \frac{2(25)-6(5)+3}{3(5)-9} = \frac{23}{6}$. So we have the point $(5, \frac{23}{6})$.

Final Answer: