Question

A bridge spans a distance of 1012m. Large cables are attached to the tops of the towers, 45 m above the road. The road is suspended from the large cables by many smaller vertical cables. The smallest vertical cable measures about 1.5m. Determine a quadratic equation for the large cables.

Answer 1

Answer:

A quadratic equation for the large cables is $f(x) = \dfrac{87 }{512072} \cdot x^2 + 1.5$

Step-by-step explanation:

The shape of the quadratic equation representing the large cables is a parabola

Taking the point of the smallest vertical cables as the vertex, (h, k) with coordinates, (0, 1.5), we have;

h = -b/(2·a)

∴ 0 = -b/(2·a), from which we have, b = 2·a × 0 = 0

k = 1.5 = c - (b²/(4·a) = c - (0/(4·a)) = c

∴ c = 1.5

The standard form of the quadratic equation, a·(x - h)² + k is therefore, given as follows;

f(x) = a·(x - 0)² + 1.5 = a·x² + 1.5

At the towers which are on either side of the bridge, when x = 1012/2 = 506, f(x) = y = 45

Therefore, we have;

45 = a·506² + 1.5

a = (45 - 1.5)/506² = 87/512072 ≈ 1.699 × 10⁻⁴

The quadratic equation for the large cables, f(x), can therefore be presented as follows;

$f(x) = \dfrac{87 }{512072} \cdot x^2 + 1.5$