Problem 60 - Algebra A Combined Function

Question

Solve. 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.

Step-by-Step Solution

Answer

The equation of the parabolic arch is \(y = -\frac{2}{125}x^2 + 40\).

Step 1: Understand the Problem

The arch of the bridge forms a parabola. The width of the arch, which is the same as the length of the road over the arch, is 100 meters. The maximum height of the arch, the vertex of the parabola, is 40 meters. We need to find the equation representing this parabolic arch.

Step 2: Determine the Vertex Form of the Parabola

We place the vertex of the parabola at the origin of our coordinate system for simplicity, making the vertex (0, 40). The equation of a parabola in vertex form is \( y = a(x - h)^2 + k \), where \( (h, k) \) is the vertex. Here \( h = 0 \) and \( k = 40 \), so the equation simplifies to \( y = a(x - 0)^2 + 40 \) or \( y = ax^2 + 40 \).

Step 3: Use the Endpoints to Find 'a'

The endpoints, where the road meets the arch, are located 50 meters on either side of the vertex (since the road is 100 meters long and centered at the vertex). So, the points \((50, 0)\) and \((-50, 0)\) lie on the parabola. Using the left endpoint \((-50, 0)\), substitute \(x = -50\) and \(y = 0\) into the equation \(y = ax^2 + 40\), resulting in \[0 = a(-50)^2 + 40\]Simplifying, \[0 = 2500a + 40\]Thus, \[2500a = -40\], and \[a = -\frac{40}{2500} = -\frac{2}{125}\].

Step 4: Write the Final Equation

Substitute \(a = -\frac{2}{125}\) back into the parabolic equation obtained in Step 2: \[y = -\frac{2}{125}x^2 + 40\].This is the equation of the parabola representing the arch of the bridge.

Key Concepts

Vertex Form of a Parabola

When dealing with parabolic equations, the vertex form is a particularly useful representation. It is expressed as \( y = a(x - h)^2 + k \), where \((h, k)\) marks the vertex of the parabola. The vertex is a significant point, representing either the maximum or minimum value of the parabola depending on whether it opens up or down.

In the context of the bridge problem, the vertex is the highest point of the arch, which was identified as \((0, 40)\). By placing the vertex at this point, we simplify the equation to \( y = ax^2 + 40 \), since \( h = 0 \) and \( k = 40 \). Here, \(a\) dictates the parabola's shape and its direction of opening.

• If \(a\) is positive, the parabola opens upwards.
• If \(a\) is negative, the parabola opens downwards.

This helps in visualizing how the arch behaves over its span.

Quadratic Functions

Quadratic functions are pivotal in understanding parabolas and their characteristics.

They are typically represented by equations of the form \( ax^2 + bx + c \). However, in the vertex form, as with the bridge, we start from \( y = ax^2 + 40 \). Here, the absence of linear term \( bx \) indicates symmetry about the y-axis given the placement of vertex at origin along the x-axis. Quadratic functions have distinctive features:

• Shape: Parabolas either arch upwards or downwards based on the sign of \(a\).
• Roots: These are the points where the parabola intersects the x-axis, often where the bridge meets the ground.
• Vertex: In our problem, it is the maximum height of the parabola, key to determining the function's peak value.

These elements make quadratic functions robust for modeling scenarios like bridges or tunnels.

Coordinate Geometry

Coordinate geometry blends algebra and geometry to study figures using a coordinate system.

In our bridge scenario, it helps understand parabolic structures in real-world contexts. By placing figures like the parabolic arch in a coordinate system, complex problems become visual and solvable with algebraic methods.A crucial aspect here is identifying points and distances:

• Vertex: Placing at \((0, 40)\) provides a reference point for building the equation.
• Distance: The 100-meter span translates to identifying endpoints \((50, 0)\) and \((-50, 0)\).
• Symmetry: The parabola's symmetric properties simplify computations, centralizing around the vertex.

By combining formulae and properties specific to parabolas, coordinate geometry becomes instrumental in forming equations that model physical structures like the bridge arch accurately.