Let's solve and analyze the rational function $f(x) = \frac{3x^2 + 6x}{x^2 + x}$ step by step.

Step 1: Simplify the Rational Function

We start by factoring the numerator and the denominator:

$f(x) = \frac{3x(x+2)}{x(x+1)}$

The $x$ in the numerator and denominator can cancel out, but only for values of $x \neq 0$. So, the simplified form is:

$f(x) = \frac{3(x+2)}{x+1}, \quad x \neq 0$

Now, we'll find the key properties of this function.

Step 2: Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero, provided that the numerator is non-zero at that point.

The denominator of the simplified function is $x+1$, so the vertical asymptote occurs at:

$x + 1 = 0 \quad \Rightarrow \quad x = -1$

Thus, there is a vertical asymptote at $x = -1$.

Step 3: Horizontal Asymptotes

To find the horizontal asymptote, we look at the behavior of the function as $x \to \infty$ or $x \to -\infty$.

The degrees of the numerator and denominator are the same (both are degree 1 after simplification). The horizontal asymptote is given by the ratio of the leading coefficients. Here, the leading coefficient in the numerator is 3 and in the denominator is 1. Therefore, the horizontal asymptote is:

$y = \frac{3}{1} = 3$

Step 4: X-Intercept

To find the x-intercept, we set the numerator equal to zero:

$3(x+2) = 0 \quad \Rightarrow \quad x + 2 = 0 \quad \Rightarrow \quad x = -2$

Thus, the x-intercept is at $(-2, 0)$.

Step 5: Y-Intercept

To find the y-intercept, we substitute $x = 0$ into the simplified function:

$f(0) = \frac{3(0+2)}{0+1} = \frac{6}{1} = 6$

Thus, the y-intercept is at $(0, 6)$.

Step 6: Tabulated Form

We can create a table of values by choosing several values of $x$ and evaluating $f(x)$. This will help visualize the graph.

$x$	$f(x)$
-3	1.5
-2	0
-1.5	1
-1.1	-3.3
-0.5	3
0	6
1	7.5
2	9

Step 7: Graphing the Function

Based on the table of values and the information we gathered (asymptotes, intercepts), we can sketch the graph. The key features are:

• Vertical asymptote at $x = -1$
• Horizontal asymptote at $y = 3$
• X-intercept at $(-2, 0)$
• Y-intercept at $(0, 6)$

The graph will approach the vertical asymptote $x = -1$ without touching it, and as $x \to \infty$ or $x \to -\infty$, the graph will approach the horizontal asymptote $y = 3$.

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Summary

• Vertical Asymptote: $x = -1$
• Horizontal Asymptote: $y = 3$
• X-Intercept: $(-2, 0)$
• Y-Intercept: $(0, 6)$
• Simplified Function: $f(x) = \frac{3(x+2)}{x+1}$

Do you have any questions or want more details on any part of this process?

5 Related Questions:

1. How do you find slant asymptotes in rational functions?
2. What happens when the degrees of the numerator and denominator are not the same?
3. How do vertical asymptotes affect the graph of a function near them?
4. Can a rational function have more than one horizontal asymptote?
5. What role do holes in the graph play for rational functions?

Tip:

Always check if you can simplify a rational function before proceeding with analyzing its properties like intercepts and asymptotes. This makes calculations much easier!