Problem 32 - Algebra 2

Question

ARCHITECTURE. For Exercises 32 and \(33,\) use the following information. The shape of each arch supporting the Exchange House can be modeled by \(h(x)=-0.025 x^{2}+2 x,\) where \(h(x)\) represents the height of the arch and \(x\) represents the horizontal distance from one end of the base in meters. Write the equation of the axis of symmetry and find the coordinates of the vertex of the graph of \(h(x) .\)

Step-by-Step Solution

Answer

Axis of symmetry: \(x = 40\); Vertex: \((40, 40)\).

Step 1: Identify the Equation Type

The given equation is a quadratic equation of the form \(h(x) = ax^2 + bx + c\). Here, \(a = -0.025\), \(b = 2\), and \(c = 0\). Since \(a < 0\), the parabola opens downwards.

Step 2: Calculate the Axis of Symmetry

The axis of symmetry for a quadratic equation \(ax^2 + bx + c\) is given by the formula \(x = -\frac{b}{2a}\). First, plug the values of \(b\) and \(a\) into the formula: \(-\frac{2}{2 \times -0.025} = -\frac{2}{-0.05} = 40\). Therefore, the equation of the axis of symmetry is \(x = 40\).

Step 3: Find the Vertex

The vertex of the parabola occurs at the axis of symmetry. Thus, the x-coordinate of the vertex is \(x = 40\). To find the y-coordinate, substitute \(x = 40\) back into the equation for \(h(x)\): \(h(40) = -0.025(40)^2 + 2(40)\). Calculate as follows: \(h(40) = -0.025 \cdot 1600 + 80 = -40 + 80 = 40\). Thus, the coordinates of the vertex are \((40, 40)\).

Key Concepts

Axis of Symmetry

In a quadratic equation like the one given, the axis of symmetry is an imaginary vertical line that divides the parabola into two mirror-image halves. For the equation \( h(x) = -0.025x^2 + 2x \), the axis of symmetry can be found using the formula:

• \( x = -\frac{b}{2a} \)

This formula comes from rearranging the components of the quadratic formula to isolate the x-component that represents symmetry.
The values given in the exercise are \( a = -0.025 \) and \( b = 2 \). Substituting these into the equation gives:

• \( x = -\frac{2}{2 \times -0.025} = 40 \)

This means the imaginary line that splits the parabola is located at \( x = 40 \). Understanding this concept is important because it reveals where the maximum or minimum point of the parabola is located.

Vertex

The vertex of a parabola is a crucial point because it represents either the highest or lowest point on the graph, depending on the orientation of the parabola. In 'h(x) = -0.025x^2 + 2x', we know the parabola opens downwards as indicated by a negative \( a \) value. Hence, the vertex is the maximum point.
The x-coordinate of the vertex is found using our axis of symmetry, which we calculated to be \( x = 40 \). To find the y-coordinate, substitute \( x = 40 \) back into the original quadratic equation:

• \( h(40) = -0.025(40)^2 + 2(40) \)

Carrying out the calculation,

• \( h(40) = -0.025 \times 1600 + 80 = -40 + 80 = 40 \)

Thus, the coordinates of the vertex are \( (40, 40) \). Recognizing the vertex coordinates is vital as they provide crucial information about the graph's extremal point.

Parabola

A parabola is a symmetrical curve that forms the graph of a quadratic function like \( h(x) \). The shape and position of the parabola are determined by its equation. Here, the equation \( h(x) = -0.025x^2 + 2x \) outlines a parabola that opens downwards due to the negative sign of \( a \).
The importance of the parabola lies in several features:

• The vertex, which here is the highest point at \( (40, 40) \).
• The axis of symmetry at \( x = 40 \), splitting the parabola into two equal halves.
• The direction it opens, indicating all other graph properties, is downward because \( a < 0 \).

Understanding the trait that it can model real-world situations, such as a bridge arch, makes recognizing the parabola's properties in algebra essential.

Graphing

Graphing a quadratic equation like \( h(x) = -0.025x^2 + 2x \) allows students to visually comprehend the algebraic relationships. Visualizing a parabola involves:

• Plotting the vertex \( (40, 40) \) on a Cartesian plane.
• Drawing the axis of symmetry parallel to the y-axis at \( x = 40 \).
• Marking additional points from the equation by selecting x-values, computing accompanying y-values, and plotting these coordinates.

The downward opening of the parabola instructs students on how to graph in the direction determined by \( a \)'s negative value.
Accurate graphing helps translate abstract algebra into visual learning. Furthermore, by recognizing patterns in the graph, such as symmetry and the extremum point, students solidify their understanding of quadratic behaviors and functions.