Answer :
To solve this problem, we need to find the quadratic equation that represents the parabola formed by the main cable of the suspension bridge. The general form of the quadratic equation is given by:
[tex]\[ y = a(x-h)^2 + k \][/tex]
Where:
- [tex]\( (h, k) \)[/tex] is the vertex of the parabola,
- [tex]\( a \)[/tex] is a constant that determines the "width" and "direction" of the parabola,
- [tex]\( x \)[/tex] is the horizontal distance from the left bridge support,
- [tex]\( y \)[/tex] is the height of the cable above the roadway.
Given data:
1. The lowest point (vertex) of the cable is at a height of [tex]\( 6 \)[/tex] feet and a horizontal distance of [tex]\( 90 \)[/tex] feet from the left bridge support. Therefore, [tex]\( h = 90 \)[/tex] and [tex]\( k = 6 \)[/tex].
2. At a horizontal distance of [tex]\( 30 \)[/tex] feet, the cable is [tex]\( 15 \)[/tex] feet above the roadway.
First, we substitute the given point [tex]\( (30, 15) \)[/tex] into the equation to find the constant [tex]\( a \)[/tex]:
[tex]\[ 15 = a(30-90)^2 + 6 \][/tex]
Solving for [tex]\( a \)[/tex]:
[tex]\[ 15 = a(-60)^2 + 6 \][/tex]
[tex]\[ 15 = 3600a + 6 \][/tex]
[tex]\[ 9 = 3600a \][/tex]
[tex]\[ a = \frac{9}{3600} = 0.0025 \][/tex]
Therefore, the equation of the parabola is:
[tex]\[ y = 0.0025(x-90)^2 + 6 \][/tex]
Next, we determine the height of the cable at the left bridge support (when [tex]\( x = 0 \)[/tex]):
[tex]\[ y = 0.0025(0 - 90)^2 + 6 \][/tex]
[tex]\[ y = 0.0025 \times 8100 + 6 \][/tex]
[tex]\[ y = 20.25 + 6 \][/tex]
[tex]\[ y = 26.25 \][/tex]
Thus, the height at which the main cable attaches to the left bridge support is [tex]\( 26.25 \)[/tex] feet.
In conclusion, the quadratic equation that models this situation correctly is:
[tex]\[ y = 0.0025(x-90)^2 + 6 \][/tex]
And the main cable attaches to the left bridge support at a height of [tex]\( 26.25 \)[/tex] feet above the roadway.
[tex]\[ y = a(x-h)^2 + k \][/tex]
Where:
- [tex]\( (h, k) \)[/tex] is the vertex of the parabola,
- [tex]\( a \)[/tex] is a constant that determines the "width" and "direction" of the parabola,
- [tex]\( x \)[/tex] is the horizontal distance from the left bridge support,
- [tex]\( y \)[/tex] is the height of the cable above the roadway.
Given data:
1. The lowest point (vertex) of the cable is at a height of [tex]\( 6 \)[/tex] feet and a horizontal distance of [tex]\( 90 \)[/tex] feet from the left bridge support. Therefore, [tex]\( h = 90 \)[/tex] and [tex]\( k = 6 \)[/tex].
2. At a horizontal distance of [tex]\( 30 \)[/tex] feet, the cable is [tex]\( 15 \)[/tex] feet above the roadway.
First, we substitute the given point [tex]\( (30, 15) \)[/tex] into the equation to find the constant [tex]\( a \)[/tex]:
[tex]\[ 15 = a(30-90)^2 + 6 \][/tex]
Solving for [tex]\( a \)[/tex]:
[tex]\[ 15 = a(-60)^2 + 6 \][/tex]
[tex]\[ 15 = 3600a + 6 \][/tex]
[tex]\[ 9 = 3600a \][/tex]
[tex]\[ a = \frac{9}{3600} = 0.0025 \][/tex]
Therefore, the equation of the parabola is:
[tex]\[ y = 0.0025(x-90)^2 + 6 \][/tex]
Next, we determine the height of the cable at the left bridge support (when [tex]\( x = 0 \)[/tex]):
[tex]\[ y = 0.0025(0 - 90)^2 + 6 \][/tex]
[tex]\[ y = 0.0025 \times 8100 + 6 \][/tex]
[tex]\[ y = 20.25 + 6 \][/tex]
[tex]\[ y = 26.25 \][/tex]
Thus, the height at which the main cable attaches to the left bridge support is [tex]\( 26.25 \)[/tex] feet.
In conclusion, the quadratic equation that models this situation correctly is:
[tex]\[ y = 0.0025(x-90)^2 + 6 \][/tex]
And the main cable attaches to the left bridge support at a height of [tex]\( 26.25 \)[/tex] feet above the roadway.
