Problem 28 - Algebra and Trigonometry

Question

A quadratic function is given. (a) Express the quadratic function in standard form. (b) Sketch its graph. (c) Find its maximum or minimum value. $$ f(x)=1-6 x-x^{2} $$

Step-by-Step Solution

Answer

Standard form: \( -x^2 - 6x + 1 \), maximum value: 10 at \( x = -3 \).

Step 1: Express the Quadratic Function in Standard Form

The given function is \( f(x) = 1 - 6x - x^2 \). The standard form of a quadratic function is \( ax^2 + bx + c \). To express the given function in standard form, rearrange the terms: \(-x^2 - 6x + 1 \). Thus, \( a = -1 \), \( b = -6 \), and \( c = 1 \).

Step 2: Determine the Vertex Form

The vertex form of a quadratic function is \( f(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex. To convert \(-x^2 - 6x + 1\) to vertex form, complete the square: start with \(-1(x^2 + 6x) + 1\). To complete the square inside the parentheses: add and subtract \((\frac{6}{2})^2 = 9\), leading to:\(-1((x^2 + 6x + 9) - 9) + 1 \). Simplify, you get \(-1((x + 3)^2 - 9) + 1 \). Distribute \(-1\): \(-1(x + 3)^2 + 9 + 1 \), which becomes \( -(x + 3)^2 + 10 \). Thus, the vertex form is \(-1(x + 3)^2 + 10 \).

Step 3: Sketch the Graph of the Function

The function \( f(x) = -(x + 3)^2 + 10 \) indicates a parabola opening downward since the coefficient \(-1\) is negative. The vertex \((h, k)\) is \((-3, 10)\). Plot this point on a graph. The axis of symmetry is \( x = -3 \). Choose additional points, such as \( x = -2 \) and \( x = -4 \), and calculate \( f(-2) = 9\) and \( f(-4) = 9\). Plot these points to get the curve.

Step 4: Find Maximum or Minimum Value

Since the parabola opens downward, it has a maximum value at its vertex. The vertex is \((-3, 10)\), so the maximum value of \( f(x) \) is \( 10 \).

Key Concepts

Standard Form

The quadratic function is often expressed in its standard form for easier analysis, which is represented as \[ax^2 + bx + c\].
This form helps identify the key coefficients that define the quadratic function, namely *a*, *b*, and *c*.

• Here, *a* is the coefficient of the squared term.
• *b* is the coefficient of the linear term.
• *c* is the constant term.

From the exercise, the function is given as\[f(x) = 1 - 6x - x^2.\]
To convert this into standard form, reorder the terms based on their degrees:\[-x^2 - 6x + 1.\]
It is in standard form with

• *a* = -1
• *b* = -6
• *c* = 1

This arrangement is crucial in further transforming the equation to other formats like the vertex form.

Vertex Form

Transforming a quadratic function into vertex form aids in easily identifying the vertex of the parabola, which is critical for graphing. The vertex form is\[f(x) = a(x - h)^2 + k,\]where

• *h* and *k* represent the coordinates of the vertex.
• *a* indicates the direction of the curve (it opens upwards if *a* is positive and downwards if negative).

To convert the expression \(-x^2 - 6x + 1\) to vertex form, we use the completing the square method. Begin by factoring out \(-1\):\[-1(x^2 + 6x) + 1.\]
Next, complete the square inside the parenthesis by adding and subtracting \((\frac{6}{2})^2 = 9\). This transforms the equation:\[-1((x^2 + 6x + 9) - 9) + 1 = -1((x + 3)^2 - 9) + 1.\]
When simplified,\[-(x + 3)^2 + 10,\]the vertex form clearly shows the vertex at

• \((-3, 10)\).

Maximum Value

The maximum or minimum value of a quadratic function, expressed in vertex form, is found directly from the vertex (h, k).
When the parabola opens downwards (coefficient *a* is negative), like in the case of \(-(x + 3)^2 + 10\), the vertex provides the maximum value.
Here, the vertex is

• \((-3, 10)\),

indicating that the maximum value of the function is 10. The coordinates also imply that this maximum is reached when \(x = -3\).
This is useful for quickly determining the peak of the parabola without graphing.

Completing the Square

Completing the square is a method used to convert a quadratic function from standard form to vertex form. The aim is to form a perfect square trinomial. This is done by altering the quadratic and linear terms slightly to reveal an underlying square pattern.
Given the quadratic function in standard form \[-x^2 - 6x + 1,\]first factor out the leading coefficient from the first two terms:\[-1(x^2 + 6x) + 1.\]
Next, take half the linear coefficient, square it, and add and subtract this value:

• Linear term coefficient: 6
• Half of it: 3
• Square it: 9

Add and subtract 9 inside the parentheses:\[-1(x^2 + 6x + 9 - 9) + 1.\]
This simplifies to\[-1((x + 3)^2 - 9) + 1.\]
Distribute the \(-1\) and simplify to get the vertex form:\[-(x + 3)^2 + 10.\]
Completing the square is thus a powerful tool for revealing the geometry of a quadratic function.