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

1. Finding the y-intercept:

The y-intercept occurs when $x = 0$. Substituting $x = 0$ into the function: $f(0) = \frac{3(0) + 4}{2(0)^2 + 3(0) + 1} = \frac{4}{1} = 4$ So, the y-intercept is $(0, 4)$.

2. Finding the x-intercept:

The x-intercept occurs when $f(x) = 0$, which happens when the numerator equals zero. Set the numerator $3x + 4 = 0$: $3x + 4 = 0 \quad \Rightarrow \quad x = -\frac{4}{3}$ Thus, the x-intercept is $\left( -\frac{4}{3}, 0 \right)$.

3. Finding the vertical asymptotes:

Vertical asymptotes occur where the denominator equals zero (and the numerator does not). Set the denominator $2x^2 + 3x + 1 = 0$: $2x^2 + 3x + 1 = 0$ We can solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 2$, $b = 3$, and $c = 1$. Substituting these values: $x = \frac{-3 \pm \sqrt{3^2 - 4(2)(1)}}{2(2)} = \frac{-3 \pm \sqrt{9 - 8}}{4} = \frac{-3 \pm \sqrt{1}}{4} = \frac{-3 \pm 1}{4}$ This gives two solutions: $x = \frac{-3 + 1}{4} = \frac{-2}{4} = -\frac{1}{2} \quad \text{and} \quad x = \frac{-3 - 1}{4} = \frac{-4}{4} = -1$ Thus, the vertical asymptotes are at $x = -\frac{1}{2}$ and $x = -1$.

4. Finding the horizontal asymptote:

To determine the horizontal asymptote, we look at the degrees of the numerator and denominator.

• The degree of the numerator $3x + 4$ is 1.
• The degree of the denominator $2x^2 + 3x + 1$ is 2.

Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is at $y = 0$.

5. Graphing the function:

The key features of the graph include:

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

Would you like me to plot the graph or provide more details on any of these steps?

Related Questions:

1. How do we find slant asymptotes for rational functions?
2. What happens when the numerator degree is higher than the denominator's?
3. How do vertical asymptotes affect the behavior of functions near them?
4. What are the key differences between asymptotes and intercepts?
5. How do you find the turning points of a rational function?

Tip: Always check the behavior near asymptotes by testing values slightly greater and less than the asymptote to see the function's direction.