The given equation is $y = \sqrt{x^2 + 2x}$.

To better understand the properties of this function, let's analyze a few aspects of it:

1. Domain:

For $y = \sqrt{x^2 + 2x}$ to be defined, the expression inside the square root must be non-negative: $x^2 + 2x \geq 0$ Factoring the quadratic expression: $x(x + 2) \geq 0$ The solution to this inequality will give the domain of the function.

• $x(x + 2) = 0$ at $x = 0$ and $x = -2$.
• The sign chart analysis for the intervals determined by these points shows that the expression is non-negative when $x \leq -2$ or $x \geq 0$.

Thus, the domain is $(-\infty, -2] \cup [0, \infty)$.

2. Range:

Since the square root function always gives non-negative results, $y \geq 0$. Therefore, the range is $[0, \infty)$.

3. Behavior at key points:

• At $x = 0$: $y = \sqrt{0^2 + 2(0)} = 0$
• At $x = -2$: $y = \sqrt{(-2)^2 + 2(-2)} = \sqrt{4 - 4} = 0$

4. Asymptotic behavior:

• As $x \to \infty$, $x^2 + 2x$ becomes large, so $y \approx x$ for large positive $x$.
• As $x \to -\infty$, the function is not defined, since the domain restricts negative values of $x$ to $x \leq -2$.

Do you want more details or have any questions?

Here are five related questions:

1. How would the function change if we altered the coefficient of $x^2$?
2. Can we find the critical points of this function?
3. What happens to the graph for values outside the domain?
4. What is the slope of the tangent at $x = 1$?
5. How would you graph this function?

Tip: When solving inequalities involving quadratic expressions, sketching a number line helps visualize where the expression is positive or negative.