Problem 6

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

Verified
Answer
The function in standard form is \( f(x) = (x-6)^2 - 4 \), and the vertex is \( (6, -4) \).
1Step 1: Identify the Standard Form
The standard form of a quadratic function is given by \( f(x) = a(x-h)^2 + k \), where \((h, k)\) is the vertex of the parabola. Our goal is to rewrite the given function in this form.
2Step 2: Start Completing the Square
Given the quadratic function \( f(x) = x^2 - 12x + 32 \), we need to complete the square for the quadratic and linear terms. First, consider \( x^2 - 12x \). Take half of the coefficient of \( x \) (which is -12), square it to complete the square, \((-12/2)^2 = 36\).
3Step 3: Rewrite the Quadratic Expression
Add and subtract 36 within the expression to balance it: \( f(x) = (x^2 - 12x + 36) - 36 + 32 \). Simplify the constants: \( f(x) = (x^2 - 12x + 36) - 4 \).
4Step 4: Factor the Perfect Square Trinomial
Rewrite the expression \( x^2 - 12x + 36 \) as a square of a binomial: \( f(x) = (x - 6)^2 - 4 \). Now the quadratic function is in standard form.
5Step 5: Identify the Vertex from the Standard Form
Using the standard form \( f(x) = (x-h)^2 + k \), the vertex can be identified as \( (h, k) \). From \( f(x) = (x - 6)^2 - 4 \), the vertex \( (h, k) \) is \( (6, -4) \).

Key Concepts

Standard FormCompleting the SquareVertex of a Parabola
Standard Form
In the world of quadratic functions, the standard form is a way to express functions that makes it easier to identify key features like the vertex. Typically, a quadratic function is first presented as a trinomial: \( ax^2 + bx + c \). However, converting it into the standard form reveals much more. The standard form of a quadratic function is expressed as: \( f(x) = a(x-h)^2 + k \), where \( (h, k) \) is the vertex of the parabola.
This format highlights the function's vertex, making graphing and analysis simpler. The value \( a \) in standard form, similar to the expanded form, determines the parabola's openness direction and width. If \( a > 0 \), the parabola opens upwards, and if \( a < 0 \), it opens downwards.
  • Using the standard form, it is a breeze to identify the vertex.
  • Completing the square is often the route taken to convert from trinomial form to standard form.
Completing the Square
Completing the square is a mathematical technique used to convert quadratic functions into a form that makes it easy to read off their key properties, like the vertex. It involves restructuring part of the quadratic expression into a perfect square trinomial.
Let's break down the process using the exercise: We have \( f(x) = x^2 - 12x + 32 \). First, focus on the quadratic and linear terms: \( x^2 - 12x \). To complete the square:
  • Take the coefficient of \( x \) (here, -12), and halve it to get -6.
  • Square this result (-6), which gives 36.
Add and subtract this square within the equation: \( f(x) = (x^2 - 12x + 36) - 36 + 32 \).
This ensures the expression remains equivalent to the original. Now, the terms inside the parenthesis form a perfect square trinomial:\( (x - 6)^2 \). Simplifying further, we get \( f(x) = (x - 6)^2 - 4 \).
Completing the square transforms the equation into one that's easy to analyze and graph.
Vertex of a Parabola
The vertex of a parabola is a crucial point, representing the minimum or maximum value of a quadratic function, depending on the parabola's orientation. Once the quadratic is in standard form, \( f(x) = a(x-h)^2 + k \), identifying the vertex becomes straightforward: it's \( (h, k) \). For our function \( f(x) = (x - 6)^2 - 4 \), the vertex is \( (6, -4) \).
Here's why the vertex is important:
  • The vertex tells us the turning point of the parabola.
  • In maximization and minimization problems, the vertex offers critical insights.
  • The x-coordinate \( h \) tells us where the vertex is located along the x-axis, while \( k \) shows its position along the y-axis, indicating the parabola's minimum or maximum value.
Thus, understanding the vertex is key to grasping the overall behavior and graph of the quadratic function.