The function given in the image is:

$y = \frac{4x^2 + 12x + 1}{x - 3}.$

Let us describe this function by analyzing its characteristics:

1. Domain

The function is undefined where the denominator $x - 3 = 0$. Therefore, the domain is: $x \in \mathbb{R}, \, x \neq 3.$

2. Asymptotes

• Vertical Asymptote: A vertical asymptote occurs where the denominator equals zero, which is at $x = 3$.
• Horizontal Asymptote: To find the horizontal asymptote, analyze the behavior of the function as $x \to \pm\infty$:
  • The highest degree term dominates both numerator and denominator. As $x \to \pm\infty$, the function behaves like $\frac{4x^2}{x} = 4x$, which suggests there is no horizontal asymptote since the degree of the numerator is higher than the denominator.

Instead, there is an oblique asymptote. Divide $4x^2 + 12x + 1$ by $x - 3$ using polynomial division to find the equation of the oblique asymptote: $\text{Division: } \frac{4x^2 + 12x + 1}{x - 3} = 4x + 24 + \frac{73}{x - 3}.$ The oblique asymptote is $y = 4x + 24$.

3. Intercepts

• x-Intercepts: Set $y = 0$, which means the numerator must be zero: $4x^2 + 12x + 1 = 0.$ Solve using the quadratic formula: $x = \frac{-12 \pm \sqrt{12^2 - 4(4)(1)}}{2(4)} = \frac{-12 \pm \sqrt{144 - 16}}{8} = \frac{-12 \pm \sqrt{128}}{8} = \frac{-12 \pm 8\sqrt{2}}{8} = -1.5 \pm \sqrt{2}.$ So the x-intercepts are $x = -1.5 + \sqrt{2}$ and $x = -1.5 - \sqrt{2}$.

• y-Intercept: Set $x = 0$: $y = \frac{4(0)^2 + 12(0) + 1}{0 - 3} = \frac{1}{-3} = -\frac{1}{3}.$ The y-intercept is $(0, -\frac{1}{3})$.

4. Behavior Around Asymptotes

Near $x = 3$, the function diverges to $\pm\infty$ (depending on whether $x$ approaches $3^+$ or $3^-$).

5. Overall Shape

The function is a rational expression with an oblique asymptote $y = 4x + 24$ and a vertical asymptote at $x = 3$. It crosses the x-axis at two points and has one y-intercept.

Would you like more details on the graph or clarification?

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Related Questions

1. How do you graph functions with oblique asymptotes?
2. What is the significance of the numerator's degree being higher than the denominator's?
3. How do vertical asymptotes affect the domain of a function?
4. How do you determine the nature of roots for quadratic equations?
5. Can a rational function have both horizontal and oblique asymptotes?

Tip

When dividing polynomials to find oblique asymptotes, only keep the linear term and ignore the remainder—it simplifies analyzing the graph!