To graph the function $f(x) = \frac{3x + 4}{2x^2 + 3x + 1}$, follow these steps:

1. Identify the asymptotes:

• Vertical Asymptotes: These occur where the denominator is zero, i.e., solve $2x^2 + 3x + 1 = 0$. Using the quadratic formula: $x = \frac{-3 \pm \sqrt{3^2 - 4(2)(1)}}{2(2)} = \frac{-3 \pm \sqrt{9 - 8}}{4} = \frac{-3 \pm 1}{4}$ So, the roots are $x = -1$ and $x = -\frac{1}{2}$. Hence, there are vertical asymptotes at $x = -1$ and $x = -\frac{1}{2}$.

• Horizontal Asymptote: Since the degree of the denominator is higher than the degree of the numerator (degree 2 in the denominator and degree 1 in the numerator), the horizontal asymptote is $y = 0$.

2. Find the intercepts:

• x-intercept: Set the numerator $3x + 4 = 0$, giving $x = -\frac{4}{3}$. So, the x-intercept is at $(-\frac{4}{3}, 0)$.
• y-intercept: Set $x = 0$ in the function: $f(0) = \frac{3(0) + 4}{2(0)^2 + 3(0) + 1} = \frac{4}{1} = 4$ So, the y-intercept is at $(0, 4)$.

3. Behavior near asymptotes and end behavior:

• As $x \to \infty$ or $x \to -\infty$, the function approaches the horizontal asymptote $y = 0$.
• Near the vertical asymptotes $x = -1$ and $x = -\frac{1}{2}$, the function will approach infinity or negative infinity, depending on the direction.

4. Plot key points:

• Vertical asymptotes at $x = -1$ and $x = -\frac{1}{2}$.
• Horizontal asymptote at $y = 0$.
• x-intercept at $(-\frac{4}{3}, 0)$ and y-intercept at $(0, 4)$.

Would you like me to generate a plot of this graph or need more details on any step?

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Here are 5 related questions:

1. How do you determine slant asymptotes for a rational function?
2. What is the significance of the degree of the numerator and denominator in determining asymptotes?
3. How do you find the local maxima and minima of rational functions?
4. What happens when the degree of the numerator is higher than the denominator?
5. How do you analyze the concavity of rational functions?

Tip: Always check both the numerator and denominator when finding intercepts and asymptotes for rational functions.