To solve for $\frac{g(x)}{h(x)}$ given the functions:

$g(x) = -x^2 - 2x$ $h(x) = -3x - 4$

we need to find the expression for $\frac{g(x)}{h(x)}$.

Solution:

1. Substitute $g(x)$ and $h(x)$ into the fraction: $\frac{g(x)}{h(x)} = \frac{-x^2 - 2x}{-3x - 4}$

2. Simplify the expression if possible.

The expression $\frac{-x^2 - 2x}{-3x - 4}$ cannot be factored further to simplify the fraction directly. Therefore, the answer remains:

$\frac{g(x)}{h(x)} = \frac{-x^2 - 2x}{-3x - 4}$

Alternatively, you could leave the negative signs outside and rewrite it as:

$\frac{g(x)}{h(x)} = \frac{x^2 + 2x}{3x + 4}$

Would you like further details or have any questions?

Related Questions

1. How do you determine the domain of $\frac{g(x)}{h(x)}$?
2. Can this expression be further simplified if specific values for $x$ are known?
3. How would you calculate $g(x) \times h(x)$ instead of $\frac{g(x)}{h(x)}$?
4. What happens if $x$ causes $h(x) = 0$ in this fraction?
5. Can you find the limit of $\frac{g(x)}{h(x)}$ as $x$ approaches infinity?

Tip

When simplifying rational expressions, always check if you can factor out terms to reduce the expression further.