The given equation is:

$y = \frac{6x^2}{x^2 - 1}$

1. Domain of the function:

The domain consists of all values of $x$ for which the denominator is not zero. We need to determine when:

$x^2 - 1 = 0$

Solving this:

$x^2 = 1 \quad \Rightarrow \quad x = \pm 1$

Thus, the function is undefined at $x = 1$ and $x = -1$. Therefore, the domain of the function is:

$x \in (-\infty, -1) \cup (-1, 1) \cup (1, \infty)$

2. Asymptotes:

• Vertical Asymptotes occur where the denominator is zero, which we already found at $x = \pm 1$. Therefore, vertical asymptotes are at $x = 1$ and $x = -1$.

• Horizontal Asymptote: For large values of $x$ (as $x \to \infty$ or $x \to -\infty$), the degree of the numerator and denominator is the same (both are degree 2), so we can find the horizontal asymptote by comparing the leading coefficients:

$\lim_{x \to \infty} y = \frac{6x^2}{x^2} = 6$

Thus, the horizontal asymptote is $y = 6$.

3. Behavior of the function:

• For $x \to \infty$: As $x$ increases, the value of $y$ approaches 6, the horizontal asymptote.
• For $x \to -\infty$: Similarly, as $x$ decreases, $y$ also approaches 6.
• At $x = 1$ and $x = -1$: The function exhibits vertical asymptotes because the denominator becomes zero at these points.

4. Derivative (Slope of the function):

If you need the rate of change of this function, we can differentiate it. Using the quotient rule:

$y' = \frac{d}{dx} \left( \frac{6x^2}{x^2 - 1} \right)$

The quotient rule states that:

$\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$

Let $f(x) = 6x^2$ and $g(x) = x^2 - 1$. Then:

$f'(x) = 12x, \quad g'(x) = 2x$

Substituting into the quotient rule:

$y' = \frac{(12x)(x^2 - 1) - (6x^2)(2x)}{(x^2 - 1)^2}$

Simplifying:

$y' = \frac{12x(x^2 - 1) - 12x^3}{(x^2 - 1)^2}$ $y' = \frac{12x^3 - 12x - 12x^3}{(x^2 - 1)^2}$ $y' = \frac{-12x}{(x^2 - 1)^2}$

This is the derivative of the function.

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Do you want further details on any of these steps or have any additional questions?

Related Questions:

1. How can we find the second derivative of this function?
2. How do you calculate the local maxima and minima of this function?
3. What is the behavior of the function near the vertical asymptotes?
4. How can we graph this function using the information above?
5. How would you solve for $x$ if $y = 0$?

Tip: Always check the domain of a function before graphing to avoid plotting undefined points.