Problem 6 - College Algebra with Corequisite Support

Question

For the following exercises, rewrite the quadratic functions in standard form and give the vertex. $$ f(x)=x^{2}-12 x+32 $$

Step-by-Step Solution

Answer

The standard form is \( f(x) = (x - 6)^2 - 4 \) with the vertex at \((6, -4)\).

Step 1: Identify the Original Form

The given quadratic function is \( f(x) = x^2 - 12x + 32 \). The coefficients are \( a = 1 \), \( b = -12 \), and \( c = 32 \).

Step 2: Complete the Square

To complete the square, first, focus on the terms \( x^2 - 12x \). Take half of the coefficient of \( x \), square it, and add and subtract this value. Half of \(-12\) is \(-6\), and squaring it gives \(36\). The expression becomes \( x^2 - 12x + 36 - 36 + 32 \).

Step 3: Simplify the Expression

The expression \( x^2 - 12x + 36 \, - 36 + 32 \) can be rewritten by factoring: \((x - 6)^2 - 4\).

Step 4: Write in Standard Form

The standard form of a quadratic function is \( f(x) = a(x - h)^2 + k \). Comparing this with \((x - 6)^2 - 4\), we get \( a = 1 \), \( h = 6 \), \( k = -4 \). Thus, the standard form is \( f(x) = (x - 6)^2 - 4 \).

Step 5: Determine the Vertex

The vertex \((h, k)\) of the quadratic function \( f(x) = a(x - h)^2 + k \) is found from the standard form. Here, the vertex is \((6, -4)\).

Key Concepts

Completing the Square

Completing the square is a method used to rewrite quadratic functions, turning them into a form that reveals important properties such as the vertex. Here's how it works: When you have a quadratic function like \( f(x) = x^2 - 12x + 32 \), the goal is to manipulate it so that part of it becomes a perfect square trinomial. This simplifies further analysis.Let's break it down:

• Focus on the \( x^2 \) and \( x \)-terms. For our example, \( x^2 - 12x \).
• Take half of the coefficient of \( x \). So, half of \( -12 \) is \( -6 \).
• Square this number: \((-6)^2 = 36\).
• Add and subtract this squared number inside the equation to balance it: \( x^2 - 12x + 36 - 36 + 32 \).

Completing the square allows the quadratic equation to be expressed in a way that easily factors into \( (x - 6)^2 \), showing a clearer structure. This method is foundational for rewriting quadratic functions into vertex form.

Vertex Form

The vertex form of a quadratic function is particularly useful because it directly displays the vertex of the parabola represented by the function. It's given by the formula \( f(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex.In our example:

• We completed the square and got the equation to look like \( (x - 6)^2 - 4 \).
• From this, it can be easily compared to the standard vertex form formula.
• This straightforward comparison reveals that the vertex \( (h, k) \) is \( (6, -4) \).

The vertex form is particularly advantageous when graphing because it provides the vertex's location at a glance, allowing students to easily visualize the parabola's shift and orientation on a coordinate plane.

Standard Form of Quadratic Equation

The standard form of a quadratic equation is typically written as \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. In this form, the focus is on the coefficients of the equation rather than the vertex, but it is the starting point when converting to other useful forms like the vertex form.For our given function \( f(x) = x^2 - 12x + 32 \):

• It begins in standard form with its coefficients clearly identified: \( a=1 \), \( b=-12 \), \( c=32 \).
• From here, we can perform operations like completing the square to eventually express it in vertex form.

While the standard form is not as straightforward for determining the vertex, it is excellent for a variety of algebraic operations and when working with systems of equations.