Problem 98 - Intermediate Algebra

Question

A bridge constructed over a bayou has a supporting arch in the shape of a parabola. Find an equation of the parabolic arch if the length of the road over the arch is 100 meters and the maximum height of the arch is 40 meters. (IMAGE CANNOT COPY)

Step-by-Step Solution

Answer

The equation of the parabolic arch is \( y = -\frac{1}{62.5}x^2 + 40 \).

Step 1: Understand the Problem

We are given a parabolic arch with a road 100 meters long, and the maximum height of the arch is 40 meters. The problem asks for the equation of this parabola.

Step 2: Establish the Coordinate System

Let's place the vertex of the parabola at the origin of a coordinate system, with the parabola opening downwards. Therefore, the vertex of the parabola is at the point (0, 40), and the road extends from x = -50 to x = 50 since the total length of the road is 100 meters.

Step 3: Use the Standard Form of a Parabola

The standard form of a parabola with vertical axis is \( y = ax^2 + bx + c \).Since the vertex is at (0, 40), the equation simplifies to \( y = ax^2 + 40 \) with no linear term \(bx\), because it's symmetric about the y-axis.

Step 4: Use Given Points to Find 'a'

We know that at the endpoints of the road, x = 50 and x = -50, the height of the arch (y) should be 0. Using the point (50, 0) in the equation:\[ 0 = a(50)^2 + 40 \]Therefore, \( 2500a = -40 \). Solve for 'a':\[ a = -\frac{40}{2500} = -\frac{1}{62.5} \]

Step 5: Write the Final Equation

The equation of the parabola, substituting the value of 'a', is:\[ y = -\frac{1}{62.5}x^2 + 40 \]This describes the parabolic arch of the bridge.

Key Concepts

Vertex Form

The vertex form of a parabolic equation is a convenient way to express the parabola when you know the vertex point. It is written as \( y = a(x-h)^2 + k \), where \((h, k)\) is the vertex of the parabola. The "\(a\)" value affects the width and direction of the parabola - if \(a\) is positive, the parabola opens upwards, and if negative, it opens downwards.
In our bridge example, the vertex is at \((0, 40)\), making the vertex form \(y = ax^2 + 40\). Here, the parabola opens downwards because the arch represents a downward curve, used to support the bridge structure. This simplified equation helps quickly find the parabola's highest point, which is crucial in applications like bridges or arches.

Coordinate System

Using a coordinate system is vital for graphing parabolas and interpreting real-world problems like the bridge arch. In this case, we strategically place the vertex, \((0, 40)\), at the origin of our coordinate system's reference. This makes it easier to visualize and solve the problem.
The parabola's symmetry is key – it extends equally from the point \(x = 0\) to \(x = +50\) and \(x = -50\), totaling a 100-meter span as given. Understanding this setup allows us to interpret the problem intuitively and apply the correct formulas to determine the parabola's equation accurately.

Standard Form of a Parabola

The standard form of a parabola is \(y = ax^2 + bx + c\). It is helpful when the parabola's orientation and other parameters are unknown. In the bridge example, symmetry about the y-axis simplifies to \( y = ax^2 + 40 \), eliminating the \( bx \) term since the parabola opens directly upwards or downwards via the y-axis.
This form is particularly useful when needing to derive additional characteristics about the parabola, such as its direction or symmetry, by further examining the 'a' parameter and vertex placement.

Solving Quadratic Equations

Solving quadratic equations involves finding the x-values where the equation equals zero, these points are also called roots or zeros of the equation. For example, in our parabolic arch, we use the equation \( 0 = ax^2 + 40 \) at the x-values corresponding to the bridge edges, \(x = 50\) and \(x = -50\).

• First, substitute the x-values into the equation.
• Next, solve for \(a\): \( 2500a = -40 \) simplifies to \( a = -\frac{1}{62.5} \).

This process highlights how we reverse-engineer quadratic equations to find key coefficients that describe the parabola's shape, such as 'a' in our parabolic bridge example. Through these calculations, we ensure the parabola fits given real-world constraints accurately.