Problem 69 - Calculus Single Variable

Question

Let \(f(x)=a x^{2}+b x+c\) with \(a \neq 0 .\) By completing the square, determine the interval(s) on which \(f\) is increasing and the interval(s) on which \(f\) is decreasing. Use the derivative of \(f\) to verify your conclusions.

Step-by-Step Solution

Answer

The function is decreasing on \((-\infty, -\frac{b}{2a})\) and increasing on \((-\frac{b}{2a}, +\infty)\).

Step 1: Express the Quadratic in Vertex Form

The function given is quadratic so we need to complete the square to express it in vertex form: \[ f(x) = a(x-h)^2 + k \] Start from \( f(x) = ax^2 + bx + c \):Factor out \(a\) from the first two terms: \[ f(x) = a(x^2 + \frac{b}{a} x) + c \]Complete the square on \(x^2 + \frac{b}{a} x\):\[ f(x) = a \left(x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 \right) + c \]This becomes:\[ f(x) = a \left((x + \frac{b}{2a})^2 - \frac{b^2}{4a^2}\right) + c \]Simplifying gives us the vertex form:\[ f(x) = a(x + \frac{b}{2a})^2 + c - \frac{b^2}{4a} \]

Step 2: Determine Intervals Based on Vertex Form

The vertex of the parabola is at \(x = -\frac{b}{2a}\).Since \(a eq 0\) and \(a > 0\) (assuming standard orientation), the parabola opens upwards. Hence, the function is decreasing to the left of the vertex and increasing to the right of the vertex.Thus, the function is:- Decreasing on \(( -\infty, -\frac{b}{2a})\).- Increasing on \((-\frac{b}{2a}, +\infty)\).

Step 3: Differentiate the Function

To verify the intervals using calculus, find the derivative of \(f(x) = ax^2 + bx + c\):\[ f'(x) = \frac{d}{dx}(ax^2 + bx + c) = 2ax + b \]

Step 4: Analyze the Sign of the Derivative

The critical point occurs where \(f'(x) = 0\):\[ 2ax + b = 0 \] Solving for \(x\):\[ x = -\frac{b}{2a} \]- For \(x < -\frac{b}{2a}\), \(f'(x) < 0\), so \(f\) is decreasing.- For \(x > -\frac{b}{2a}\), \(f'(x) > 0\), so \(f\) is increasing.

Key Concepts

Vertex Form

The vertex form of a quadratic function is a great way to gain insights into the graph of the quadratic. It is expressed as \( f(x) = a(x-h)^2 + k \), where \( (h, k) \) is the vertex of the parabola. This form makes it easy to identify the vertex and understand the parabola's direction. A positive \( a \) means the parabola opens upwards, while a negative \( a \) means it opens downwards. In our example, we start with the standard quadratic form \( f(x) = ax^2 + bx + c \). By completing the square, we transform this to vertex form, helping us pinpoint essential aspects like the vertex and direction of the parabola.

Completing the Square

Completing the square is a crucial algebraic process used to rewrite a quadratic function into vertex form. This process involves manipulating the equation to make it easier to understand its geometric properties. For the given function \( f(x) = ax^2 + bx + c \), the steps to complete the square are:

• Factor out \( a \) from the terms involving \( x \) to get \( a(x^2 + \frac{b}{a}x) + c \).
• To create a perfect square trinomial, add and subtract \( \left(\frac{b}{2a}\right)^2 \) inside the parentheses.
• Simplify to obtain \( a \left((x + \frac{b}{2a})^2 - \frac{b^2}{4a^2} \right) + c \).
• Finally, express the equation as \( f(x) = a(x + \frac{b}{2a})^2 + c - \frac{b^2}{4a} \).

This reveals the vertex \( (-\frac{b}{2a}, c - \frac{b^2}{4a}) \), giving us valuable information about the graph's key features.

Derivative Analysis

Analyzing the derivative is an essential part of understanding a quadratic function's behavior. The derivative of a function provides information about its rate of change and the nature of its slope. For a quadratic function \( f(x) = ax^2 + bx + c \), the first derivative is \( f'(x) = 2ax + b \). This linear function tells us where the slope of the original function is zero, positive, or negative.

• The critical point where \( f'(x) = 0 \) is at \( x = -\frac{b}{2a} \).
• To the left of this point, \( f'(x) < 0 \), indicating that the function is decreasing.
• To the right, \( f'(x) > 0 \), meaning the function is increasing.

This derivative analysis confirms the conclusions drawn from the vertex form, providing a complete picture of the function's behavior.

Intervals of Increase and Decrease

Determining the intervals on which a quadratic function increases or decreases is key to understanding its graph. The process involves using both vertex form and derivative analysis. Firstly, from the vertex form \( f(x) = a(x-h)^2 + k \), the vertex \( x = -\frac{b}{2a} \) divides the graph into two intervals:

• The function decreases on \( (-\infty, -\frac{b}{2a}) \).
• It increases on \( (-\frac{b}{2a}, +\infty) \).

Secondly, the derivative \( f'(x) = 2ax + b \) confirms this, as \( f'(x) < 0 \) when \( x < -\frac{b}{2a} \) and \( f'(x) > 0 \) when \( x > -\frac{b}{2a} \). This dual approach, both algebraic and calculus-based, assures a thorough understanding of the function's increasing and decreasing characteristics.