Problem 27 - Algebra and Trigonometry

Question

19-28 \(=\) A quadratic function is given. (a) Express the quadratic function in standard form. (b) Sketch its graph. (c) Find its maximum or minimum value. $$ h(x)=1-x-x^{2} $$

Step-by-Step Solution

Answer

The quadratic function in standard form is \(-x^2 - x + 1\). Its maximum value is \(\frac{1}{4}\) at \(x = \frac{1}{2}\).

Step 1: Express in Standard Form

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

Step 2: Determine the Vertex Form

To sketch the graph, it is helpful to convert the function to vertex form \( a(x-h)^2 + k \). Since \( a = -1 \), the axis of symmetry is \( x = -\frac{b}{2a} = \frac{1}{2} \). Calculate \( h(\frac{1}{2}) \) to find the y-value of the vertex: \[ h\left(\frac{1}{2}\right) = -\left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) + 1 = -\frac{1}{4} - \frac{1}{2} + 1 = \frac{1}{4} \]. Thus, the vertex is \( \left(\frac{1}{2}, \frac{1}{4}\right) \).

Step 3: Sketch the Graph

The quadratic function \( h(x) = -x^2 - x + 1 \) is a downward-opening parabola because \( a = -1 \). The vertex of the parabola, found in Step 2, is at \( (\frac{1}{2}, \frac{1}{4}) \). Plot the vertex and additional points such as \((0, 1)\) and \((-1, 1)\) to sketch the parabola. Draw a symmetric curve around the axis of symmetry \( x = \frac{1}{2} \).

Step 4: Find the Minimum or Maximum Value

Since the parabola opens downwards, the vertex represents the maximum value. The maximum value is the y-value of the vertex, which is \( \frac{1}{4} \). Thus, the maximum value of the function is \( \frac{1}{4} \) at \( x = \frac{1}{2} \).

Key Concepts

Standard Form

The standard form of a quadratic function is represented as \( ax^2 + bx + c \). This is the conventional way of writing quadratics, making it easier to identify the coefficients of the polynomial. Here's a quick breakdown of what each component represents:

• \( a \): The coefficient of \( x^2 \), which determines the direction the parabola opens (upward if positive, downward if negative).
• \( b \): The coefficient of \( x \), influencing the position of the axis of symmetry.
• \( c \): The constant term, which indicates where the parabola will intersect the y-axis.

For the given function \( h(x) = 1 - x - x^2 \), rearranging gives us \( -x^2 - x + 1 \), already in the standard form. The coefficients are \( a = -1 \), \( b = -1 \), and \( c = 1 \). Recognizing these can help lay the foundation for understanding the parabola's properties.

Vertex Form

To understand the vertex form, it’s expressed as \( a(x-h)^2 + k \). This format makes it easy to identify the vertex of the parabola, \( (h, k) \), which is the highest or lowest point, depending on the direction the parabola opens.
To convert from standard to vertex form:

• The axis of symmetry is found using \( x = -\frac{b}{2a} \), which helps locate the x-coordinate of the vertex.
• Substitute this x-value into the original function to find the corresponding y-value.

For the function \( -x^2 - x + 1 \), the axis of symmetry is at \( x = \frac{1}{2} \). Substituting \( x = \frac{1}{2} \) into the quadratic gives \( h\left(\frac{1}{2}\right) = \frac{1}{4} \), making the vertex \( \left(\frac{1}{2}, \frac{1}{4}\right) \). This transformation to vertex form clarifies the location and nature of the vertex.

Maximum or Minimum Value

The maximum or minimum value of a quadratic function is determined by the vertex.
For functions of the form \( ax^2 + bx + c \):

• If \( a > 0 \), the parabola opens upwards, and the vertex gives the minimum value.
• If \( a < 0 \), the parabola opens downwards, and the vertex provides the maximum value.

Given that \( h(x) = -x^2 - x + 1 \) opens downward (since \( a = -1 \)), the vertex at \( \left(\frac{1}{2}, \frac{1}{4}\right) \) signifies the highest point on the graph. This means the maximum value of the function is \( \frac{1}{4} \) at \( x = \frac{1}{2} \).

Parabola Graph

The graph of a quadratic function is a U-shaped curve called a parabola. Understanding how to sketch this graph involves:

• Locating the vertex, which is the turning point and determines the maximum or minimum value.
• Identifying the axis of symmetry, which is a vertical line passing through the vertex, ensuring the graph is symmetrical on both sides.
• Plotting additional points on either side of the vertex to define the shape.

For \( h(x) = -x^2 - x + 1 \), the graph is a downward-opening parabola with the vertex at \( \left(\frac{1}{2}, \frac{1}{4}\right) \).
The axis of symmetry is \( x = \frac{1}{2} \). This provides a structure for plotting critical points and sketching a symmetric curve. Points such as \( (0, 1) \) and \( (-1, 1) \) further detail the graph on either side of the vertex, completing the parabola's shape.