Let's go through each of the problems you uploaded step by step:

7. End Behavior for the Function:

$f(x) = \frac{3x^2 + 5x + 6}{x^2 - 4}$

To determine the end behavior as $x \to +\infty$ and $x \to -\infty$, we focus on the highest-degree terms in the numerator and denominator.

• As $x \to \infty$: $f(x) \approx \frac{3x^2}{x^2} = 3$ Therefore, $\lim_{x \to +\infty} f(x) = 3$.

• As $x \to -\infty$: The same leading terms dominate: $f(x) \approx \frac{3x^2}{x^2} = 3$ Therefore, $\lim_{x \to -\infty} f(x) = 3$.

Thus, the end behavior of this function tends to 3 as $x \to \pm \infty$.

8. End Behavior for the Function:

$f(x) = \frac{10 + 5x^2 - 3x^3}{2x^3 - 4x + 12}$

Again, we'll focus on the highest-degree terms in both the numerator and the denominator.

• As $x \to +\infty$: The highest degree term in both the numerator and denominator is $-3x^3$ and $2x^3$ respectively: $f(x) \approx \frac{-3x^3}{2x^3} = -\frac{3}{2}$ So, $\lim_{x \to +\infty} f(x) = -\frac{3}{2}$.

• As $x \to -\infty$: The leading terms still dominate, so the end behavior is the same: $f(x) \approx \frac{-3x^3}{2x^3} = -\frac{3}{2}$ So, $\lim_{x \to -\infty} f(x) = -\frac{3}{2}$.

Therefore, the function tends to $-\frac{3}{2}$ as $x \to \pm \infty$.

10. End Behavior for the Function:

$f(x) = e^x$

For exponential functions like $e^x$:

• As $x \to +\infty$, $f(x) = e^x \to \infty$.
• As $x \to -\infty$, $f(x) = e^x \to 0$.

Therefore:

• $\lim_{x \to +\infty} e^x = \infty$
• $\lim_{x \to -\infty} e^x = 0$.

62. Horizontal and Vertical Asymptotes for the Rational Function:

$f(x) = \frac{5x - 2}{2x + 3}$

Horizontal Asymptotes: We check the degrees of the numerator and denominator.

• Since both the numerator and denominator are linear (degree 1), the horizontal asymptote is the ratio of the leading coefficients: $\lim_{x \to \pm \infty} f(x) = \frac{5}{2}$ Therefore, the horizontal asymptote is $y = \frac{5}{2}$.

Vertical Asymptotes: We find vertical asymptotes by setting the denominator equal to zero: $2x + 3 = 0 \quad \Rightarrow \quad x = -\frac{3}{2}$ Thus, the vertical asymptote is $x = -\frac{3}{2}$.

Would you like any further details or have any questions on these?

Here are 5 additional related questions:

1. What is the difference between horizontal and oblique asymptotes?
2. How do we find the slant asymptote of a rational function?
3. What happens to the end behavior of a rational function when the degree of the numerator is greater than that of the denominator?
4. How can we confirm vertical asymptotes using limits?
5. Why does $e^x$ tend to zero as $x \to -\infty$?

Tip: When dealing with rational functions, always check if the degree of the numerator is greater than the degree of the denominator to identify whether there is a slant or horizontal asymptote.