Question 27 - of 30 Step 1 of 1 Find the open interval(s) where the following function is increasing. decreasing or constant, Express your answer in interval notation. \[ g(x)=-(x-3)^{2}-2 \]
Solution
Step 1 ：The function given is \(g(x)=-(x-3)^{2}-2\), which is a quadratic function. The coefficient of the \(x^2\) term is negative, indicating that the graph of the function is a parabola that opens downwards.
Step 2 ：The function increases on the interval from negative infinity to the vertex and decreases from the vertex to positive infinity. The vertex of the parabola is the point \((h, k)\), where \(h\) is the value of \(x\) that makes the derivative of the function equal to zero.
Step 3 ：In this case, \(h = 3\). Therefore, the function increases on the interval \((-∞, 3)\) and decreases on the interval \((3, ∞)\).
Step 4 ：The function is never constant because it is a quadratic function, which means it always changes its value as \(x\) changes.
Step 5 ：\(\boxed{\text{Final Answer: The function } g(x) \text{ is increasing on the interval } (-\infty, 3) \text{ and decreasing on the interval } (3, \infty). \text{ The function is never constant.}}\)
