Problem 33 - A Graphical Approach to Precalculus with Limits

Question

Graph each finction in the standand viewing window of your calculator, and trace from left to right along a representative portion of the specified interval. Then fill in the blank of the following sentence with either increasing or decreasing. $$f(x)=2-x^{2} ;(-\infty, 0)$$

Step-by-Step Solution

Answer

The function is increasing on the interval \((-\infty, 0)\).

Step 1: Understand the Function

The given function is \( f(x) = 2 - x^2 \). This is a quadratic function, and its graph is a parabola. Since the coefficient of \( x^2 \) is negative, the parabola opens downwards.

Step 2: Determine Key Features of the Parabola

The vertex of the parabola is at \( (0, 2) \), since the equation is in the form \( f(x) = a(x-h)^2 + k \) with \( a = -1 \), \( h = 0 \), and \( k = 2 \). This is the maximum point of the function since the parabola opens downwards.

Step 3: Focus on the Specified Interval

The problem asks us to consider the interval \( (-\infty, 0) \). This is the left half of the parabola moving towards the vertex at \( x = 0 \).

Step 4: Analyze the Behavior on the Interval

As \( x \to -\infty \), the term \(-x^2\) becomes more negative, which makes \( f(x) = 2 - x^2 \) decrease. Thus, from left to right, as \( x \) approaches 0, the values of \( f(x) \) continuously increase.

Step 5: Complete the Sentence

Given the behavior of \( f(x) \) over the interval \( (-\infty, 0) \), we fill in the blank with the word 'increasing'. The function is increasing as \( x \) moves from \(-\infty\) towards 0.

Key Concepts

Parabola

A parabola is a specific type of graph that represents a quadratic function. It looks like a U-shaped curve, but it can also be inverted, forming an upside-down U. This shape is determined by the square term in the quadratic equation. In our example, the function \( f(x) = 2 - x^2 \) produces a parabola.

• The graph of the function \( f(x) = 2 - x^2 \) is called a parabola.
• Since the coefficient of \( x^2 \) is negative, namely \( -1 \), the parabola opens downward.

This downward opening creates a maximum point on the graph, which is the highest point that the parabola reaches.
Knowing if a parabola opens upward or downward is important because it helps determine whether its vertex is a maximum or minimum point. The vertex inverts the direction of the graph, signifying the switch from increasing to decreasing or vice versa.

Vertex

The vertex is a key feature of a parabola and is either its highest or lowest point. For the function \( f(x) = 2 - x^2 \), we identify the vertex from the equation rewritten in the form \( f(x) = a(x-h)^2 + k \).

• Here, \( a = -1 \), \( h = 0 \), and \( k = 2 \).
• The vertex is at the coordinate \((0, 2)\).
• This means the highest point on this particular parabola's graph is at \( y = 2 \) when \( x = 0 \).

Because the parabola opens downward, the vertex represents the maximum point. Understanding the vertex helps in determining the function's behavior around that point, as it sets the stage for whether the function increases or decreases as it moves away from the vertex.

Interval Analysis

Interval analysis involves examining how a function behaves over a specific range of inputs, or \( x \) values. For the quadratic function \( f(x) = 2 - x^2 \), we're interested in the interval \( (-\infty, 0) \).

• The interval \((-\infty, 0)\) stops just at the vertex \( x = 0 \).
• On this interval, as \( x \) decreases (toward \(-\infty\)), the term \(-x^2\) decreases more negatively, so \( f(x) = 2 - x^2 \) decreases.
• Consequently, as \( x \) approaches zero from the left, \( f(x) \) increases. In simple terms, any value of \( x \) closer to zero results in a higher \( f(x) \).

This analysis allows us to complete statements about the function's behavior over an interval, such as determining whether it is increasing or decreasing. For \( (-\infty, 0) \), the function \( f(x) \) is increasing as \( x \) moves from left to right towards zero.