The main cable of a suspension bridge forms a parabola, described by the equation [tex] y = a(x-h)^2 + k [/tex], where [tex] y [/tex] is the height in feet of the cable above the roadway, [tex] x [/tex] is the horizontal distance in feet from the left bridge support, [tex] a [/tex] is a constant, and [tex] (h, k) [/tex] is the vertex of the parabola.

At a horizontal distance of [tex] 30 \text{ ft} [/tex], the cable is [tex] 15 \text{ ft} [/tex] above the roadway. The lowest point of the cable is [tex] 6 \text{ ft} [/tex] above the roadway and is a horizontal distance of [tex] 90 \text{ ft} [/tex] from the left bridge support.

Which quadratic equation models the situation correctly?



Answer :

To model the situation described by the quadratic equation [tex]\(y = a(x-h)^2 + k\)[/tex], we need to determine the values of [tex]\(a\)[/tex], [tex]\(h\)[/tex], and [tex]\(k\)[/tex]. We are provided with the following information:

- The lowest point (vertex) of the cable is [tex]\(6\)[/tex] feet above the roadway and is [tex]\(90\)[/tex] feet horizontally from the left bridge support. This gives us the vertex [tex]\((h, k)\)[/tex], where [tex]\(h = 90\)[/tex] and [tex]\(k = 6\)[/tex].
- At a horizontal distance of [tex]\(30\)[/tex] feet, the height of the cable is [tex]\(15\)[/tex] feet above the roadway.

Given these values, the vertex form of our equation is:
[tex]\[ y = a(x - 90)^2 + 6 \][/tex]

Next, we use the information that at [tex]\(x = 30\)[/tex], [tex]\(y = 15\)[/tex] to find the value of [tex]\(a\)[/tex]. Substitute [tex]\(x = 30\)[/tex] and [tex]\(y = 15\)[/tex] into the equation:

[tex]\[ 15 = a(30 - 90)^2 + 6 \][/tex]

This simplifies to:
[tex]\[ 15 = a(-60)^2 + 6 \][/tex]
[tex]\[ 15 = 3600a + 6 \][/tex]
To isolate [tex]\(a\)[/tex], subtract [tex]\(6\)[/tex] from both sides:
[tex]\[ 9 = 3600a \][/tex]
Finally, solve for [tex]\(a\)[/tex]:
[tex]\[ a = \frac{9}{3600} = \frac{1}{400} \][/tex]

So, the quadratic equation that models the situation is:
[tex]\[ y = \frac{1}{400}(x - 90)^2 + 6 \][/tex]

Therefore, the correctly modeled quadratic equation is:
[tex]\[ y = \frac{1}{400}(x - 90)^2 + 6 \][/tex]