Find the Domain of (x+10)²+2: Analyzing Positive and Negative Inputs

The function given is $y = (x+10)^2 + 2$. This is a quadratic function in vertex form.

The vertex form of a quadratic function is $y = a(x-h)^2 + k$, where the vertex is $(h, k)$. For our function, $h = -10$ and $k = 2$, so the vertex is $(-10, 2)$.

In this case, since the coefficient of $(x+10)^2$ is positive (implicitly 1), the parabola opens upwards. This means the function has a minimum point at the vertex, and $y$ will only increase from that point.

Given that the vertex point has a $y$-value of 2, which is positive, the entire domain yields values of $y$ that are greater than 2. Therefore, $y$ will never be negative.

Now, let's determine the domains:

• For the negative domain, we seek values of $x$ where the function $y$ is negative. Since the minimum $y$-value is 2, no such $x$ satisfies $y \lt 0$. Hence, there is no negative domain.
• Similarly, all values of $x$ yield a positive range of $y$, as $y \geq 2$ for all $x$. Thus, for the positive domain, it is all $x$, as every $y$ value is positive or zero.

Consequently, the specified positive domain is all $x$, and the negative domain is none.

Thus, the correct answer is:

$x < 0 :$ None

$x > 0 :$ All $x$