Problem 11 - College Algebra

Question

For the following exercises, rewrite the quadratic functions in standard form and give the vertex. $$ k(x)=3 x^{2}-6 x-9 $$

Step-by-Step Solution

Answer

The standard form is \(3(x-1)^2 - 12\); vertex: \((1, -12)\).

Step 1: Identify the Quadratic Function

The given quadratic function is \(k(x) = 3x^2 - 6x - 9\). Our goal is to rewrite this in standard form, \(k(x) = a(x-h)^2 + k\), and identify \(h\) and \(k\) to find the vertex.

Step 2: Factor Out the Leading Coefficient

Factor out the leading coefficient, \(3\), from the quadratic and linear terms:\[k(x) = 3(x^2 - 2x) - 9\]

Step 3: Complete the Square

To complete the square inside the parentheses, take half of the coefficient of \(x\) inside the parenthesis, square it, and add and subtract this square inside the parentheses:The coefficient of \(x\) is \(-2\). Half of \(-2\) is \(-1\), and squaring \(-1\) gives \(1\).Add and subtract \(1\) inside the parentheses:\[k(x) = 3(x^2 - 2x + 1 - 1) - 9 = 3((x - 1)^2 - 1) - 9\]

Step 4: Simplify the Expression

Use distributive property to simplify:\[k(x) = 3(x-1)^2 - 3 - 9\]Combine like terms:\[k(x) = 3(x-1)^2 - 12\]

Step 5: Identify the Vertex

Now the quadratic is in the standard form \(k(x) = a(x-h)^2 + k\).Identify \(h = 1\) and \(k = -12\), so the vertex is \((1, -12)\).

Key Concepts

Standard Form

When working with quadratic functions, understanding the standard form can be incredibly useful. The standard form of a quadratic function is expressed as:

• \( ax^2 + bx + c \)

In this form:

• \( a \) is the coefficient of \( x^2 \), the quadratic term.
• \( b \) is the coefficient of \( x \), the linear term.
• \( c \) is the constant term.

Understanding this helps us see how the graph of the function relates to the equation itself. Each part of the quadratic equation influences the shape and position of the parabola on the graph. For example, the coefficient \( a \) determines if the parabola opens upwards or downwards. If \( a > 0 \), it opens upwards, while if \( a < 0 \), it opens downwards.

In our specific problem, the function is given as \( k(x) = 3x^2 - 6x - 9 \). This is already in standard form with \( a = 3 \), \( b = -6 \), and \( c = -9 \), making it easier to work with when converting to vertex form.

Vertex Form

The vertex form of a quadratic function is a useful way to express the equation, especially when determining the maximum or minimum points on the graph. The vertex form is written as:

• \( a(x-h)^2 + k \)

In this form:

• The vertex of the parabola is located at \((h, k)\).
• \( a \) still influences the direction the parabola opens, just as in standard form.

To convert from the standard form to the vertex form, one would use the process called "completing the square." This technique allows you to identify the \( h \) and \( k \) values, which represent the vertex coordinates. In our exercise, upon completing the square, we convert the given equation to \( k(x) = 3(x-1)^2 - 12 \). This tells us that the vertex of the parabola for our function is located at \((1, -12)\). This is crucial, as it helps to understand the parabola's position and identify features such as its axis of symmetry and vertex.

Completing the Square

Completing the square is a method for transforming a quadratic equation from standard form into vertex form. This process involves creating a perfect square trinomial, which can then be easily factored to determine the vertex.

To complete the square, follow these steps:

• First, if necessary, factor out the leading coefficient from the quadratic and linear terms.
• Next, take half the coefficient of \( x \), square it, and then add and subtract this value inside the equation.
• Finally, rearrange the equation to incorporate these changes, resulting in a perfect square trinomial.

In our original exercise \( k(x) = 3x^2 - 6x - 9 \), we took the following steps:

• Factored out a \( 3 \), resulting in \( k(x) = 3(x^2 - 2x) - 9 \).
• Calculated \( (-2)/2 = -1 \) and then squared it to get \( 1 \), adding and subtracting \( 1 \) inside the parentheses.
• This gave us \( k(x) = 3((x-1)^2 - 1) - 9 \), which simplifies to \( k(x) = 3(x-1)^2 - 12 \).

Through completing the square, we have converted the quadratic equation into vertex form, making it easier to identify critical features such as the vertex.