From the uploaded image, the problem involves analyzing the rational function $f(x) = \frac{2x^2 + x - 18}{x^2 - 9}$ to find intercepts and asymptotes.

Here’s the solution explained step-by-step:

Step 1: Simplify the function

• The denominator factors as $x^2 - 9 = (x - 3)(x + 3)$.
• The numerator factors as $2x^2 + x - 18 = (2x - 9)(x + 2)$.

So, the simplified function is: $f(x) = \frac{(2x - 9)(x + 2)}{(x - 3)(x + 3)}$

Step 2: Find the x-intercepts

• $x$-intercepts occur when the numerator equals zero, i.e., $(2x - 9)(x + 2) = 0$.
  • $2x - 9 = 0$ gives $x = \frac{9}{2} = 4.5$.
  • $x + 2 = 0$ gives $x = -2$.

So, the $x$-intercepts are $\left(-2, 0\right)$ and $\left(4.5, 0\right)$.

Step 3: Find the y-intercept

• $y$-intercept occurs when $x = 0$: $f(0) = \frac{2(0)^2 + (0) - 18}{(0)^2 - 9} = \frac{-18}{-9} = 2$

So, the $y$-intercept is $(0, 2)$.

Step 4: Find the vertical asymptotes

• Vertical asymptotes occur when the denominator equals zero, but the numerator is non-zero:
  • Denominator $x^2 - 9 = 0$ gives $x = 3$ and $x = -3$.

Thus, the vertical asymptotes are at $x = 3$ and $x = -3$.

Step 5: Find the horizontal asymptotes

• Horizontal asymptotes are determined by the degrees of the numerator and denominator:
  • Both numerator and denominator have degree 2.
  • The horizontal asymptote is the ratio of the leading coefficients: $y = \frac{2}{1} = 2$

So, the horizontal asymptote is $y = 2$.

Step 6: Check for slant asymptotes

• Slant asymptotes occur if the degree of the numerator is exactly one more than the degree of the denominator.
• Here, both numerator and denominator have the same degree, so there is no slant asymptote.

Summary of Results:

• x-intercepts: $\left(-2, 0\right), \left(\frac{9}{2}, 0\right)$
• y-intercept: $\left(0, 2\right)$
• Vertical asymptotes: $x = -3, x = 3$
• Horizontal asymptote: $y = 2$
• Slant asymptote: None (Ø).

Let me know if you'd like me to clarify or expand on any part!

Here are 5 related questions for deeper understanding:

1. How would the analysis change if the degrees of numerator and denominator were different?
2. Can there be cases where the rational function has no vertical asymptotes?
3. How do multiplicities of factors in the numerator or denominator affect the graph?
4. What happens when a factor cancels out between the numerator and denominator?
5. How do we use this analysis to sketch the graph of the function?

Tip: Always check for domain restrictions when analyzing rational functions, especially removable discontinuities!