Let's graph the rational function $f(x) = \frac{3x^2 + 6x}{x^2 + x}$.

Step-by-step explanation:

1. Simplify the Rational Function: First, simplify the expression if possible. $f(x) = \frac{3x^2 + 6x}{x^2 + x}$ Factor both the numerator and the denominator:

   • Numerator: $3x^2 + 6x = 3x(x + 2)$
   • Denominator: $x^2 + x = x(x + 1)$

   So, the simplified function is: $f(x) = \frac{3x(x + 2)}{x(x + 1)}$

   Cancel the common factor $x$ (but keep in mind that $x = 0$ is a restricted value, which we'll deal with later): $f(x) = \frac{3(x + 2)}{x + 1}, \quad x \neq 0$

2. Determine Vertical Asymptotes: A vertical asymptote occurs where the denominator is zero and the numerator is not zero. In the simplified form $f(x) = \frac{3(x + 2)}{x + 1}$, the denominator is zero when $x + 1 = 0$, which gives: $x = -1$ So, there is a vertical asymptote at $x = -1$.

3. Identify Holes: Holes occur when a factor is canceled out, but it originally caused both the numerator and denominator to be zero. In this case, the factor $x$ was canceled out, so there is a hole at $x = 0$. To find the y-coordinate of the hole, substitute $x = 0$ into the simplified function: $f(0) = \frac{3(0 + 2)}{0 + 1} = \frac{6}{1} = 6$ So, there is a hole at $(0, 6)$.

4. Find the Horizontal Asymptote: To determine the horizontal asymptote, compare the degrees of the numerator and denominator.

   • The degree of the numerator is 2.
   • The degree of the denominator is also 2.

   Since the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients of the numerator and denominator. In this case, both the numerator and denominator have leading coefficients of 3 and 1, respectively, so the horizontal asymptote is: $y = \frac{3}{1} = 3$

5. Plot Key Points and Asymptotes:

   • Vertical asymptote at $x = -1$.
   • Horizontal asymptote at $y = 3$.
   • Hole at $(0, 6)$.
   • Evaluate $f(x)$ at a few points to get more points on the graph:
     • $f(1) = \frac{3(1 + 2)}{1 + 1} = \frac{9}{2} = 4.5$
     • $f(-2) = \frac{3(-2 + 2)}{-2 + 1} = 0$

Let's graph the function now.