Question

Find the minimum value or maximum value of the function. Then describe where the function is increasing and decreasing. (Section 2.2) \(h(x)=\frac{1}{2}(x+2)^2-1\)

Short answer

The minimum value of the function \(h(x)=\frac{1}{2}(x+2)^2-1\) is at (-2, -1). The function is decreasing for \(x< -2\) and increasing for \(x > -2\).

Step-by-step solution

Identify the vertex of the quadratic function

From the equation, the vertex is (-2, -1) since the equation is in the form \(f(x) = a(x-h)^2 + k\), where (h, k) represents the vertex of the parabola.

Determine if the function is having a maximum or minimum.

Since the coefficient a in our function is positive, we know that the parabola opens upwards. Hence, the vertex (-2, -1) is the minimum of the function.

Find where the function is increasing or decreasing.

Given that the nature is parabolic and opens upwards, the function decreases until it reaches the vertex at x=-2, then it starts increasing. Therefore, the function is decreasing for \(x< -2\) and increasing for \(x > -2\).

Key concepts

Vertex of a Parabola

When staring down the path of quadratic functions, it's like embarking on a journey to mountaintop or valley floor; the vertex of a parabola is your destination. It is the highest or lowest point on the graph of a quadratic equation and gives us crucial information about the function's shape and direction.

Picture the quadratic function in the form of \(f(x) = a(x-h)^2 + k\), where \((h, k)\) is the vertex. This pinnacle point is a beacon for the rest of the graph—it dictates whether we're scaling a peak or delving into a trough. In our exercise, the vertex comes from rearranging the function into that form, revealing \((-2, -1)\) as our crucial coordinates. The sign of \(a\) indicates if the parabola opens upward or downward—positive for skyward, negative for groundward—helping us understand the context around the vertex and prepare for the journey ahead.

Maximum and Minimum Values

Deciphering the highs and lows of a quadratic function similar to watching the rise and fall of the tides—it's all about maximum and minimum values. When a parabola opens upwards like a U-shaped cup, the lowest point, the vertex, is cradling the smallest value of the function. Conversely, when it opens downwards, the vertex stands at the summit, bearing the highest value.

It's like a treasure hunt for the most or the least, and the coefficient \(a\) in \(f(x) = a(x-h)^2 + k\) sets the rules. Positive \(a\)? You've found a minimum. Negative \(a\)? That's a maximum. In the function \(h(x)=\frac{1}{2}(x+2)^2-1\), the positive coefficient whispers the secret - this function treasures a minimum at \(-1\) hidden at the vertex \(-2, -1\). This precious point tells us about the function’s lowest value and is instrumental for solving real-world problems where minimums or maximums are pivotal.

Increasing and Decreasing Intervals

The journey of a quadratic function isn't uniform; it has its ebb and flow reflected in increasing and decreasing intervals. These intervals signal the function's growth and decline as the x-values march from left to right across the graph. The vertex acts as a milestone dividing these intervals.

In a U-shaped parabola, the function is playing a game of descent before the vertex, and ascent afterward. Specifically, for our exercise function \(h(x)\), since the parabola opens upwards, it's decreasing as we approach the vertex at \(x = -2\) from the left. Like a hiker reaching the bottom of a valley, once we pass the vertex, we start to climb up, marking the beginning of the increasing interval. Therefore, the function is decreasing for \(x < -2\) and increasing for \(x > -2\) — a critical understanding for predicting the behavior of the function in different scenarios, such as optimizing profits or minimizing costs in business applications.