identify the graph of the function ( f(x)=\frac{1}{2}sqrt{x + 2}). determine when the function is positive…

Explanation:

Step1: Determine domain

For $y = \frac{1}{2}\sqrt{x + 2}$, the expression under the square - root must be non - negative. So $x+2\geq0$, which gives $x\geq - 2$.

Step2: Find when the function is positive

Since $\sqrt{x + 2}\geq0$ for $x\geq - 2$ and the coefficient $\frac{1}{2}>0$, the function $y=\frac{1}{2}\sqrt{x + 2}\geq0$ for $x\geq - 2$. It is positive for $x>-2$ because when $x = - 2$, $y = 0$.

Step3: Find when the function is negative

The function $y=\frac{1}{2}\sqrt{x + 2}$ is non - negative for its domain $x\geq - 2$, so there are no real numbers for which it is negative.

Step4: Analyze increasing and decreasing

The square - root function $y=\sqrt{u}$ is an increasing function and $u=x + 2$ is also an increasing function. The coefficient $\frac{1}{2}>0$, so the function $y=\frac{1}{2}\sqrt{x + 2}$ is increasing for $x\geq - 2$. It is not decreasing for any part of its domain.

Step5: Determine end - behavior

As $x\to\infty$, $\sqrt{x + 2}\to\infty$, so $y=\frac{1}{2}\sqrt{x + 2}\to\infty$.

Answer:

Positive: $x > - 2$ Negative: no real numbers Increasing: $x\geq - 2$ Decreasing: no real numbers End behavior: $y\to\infty$ as $x\to\infty$