Question

the cable of a suspension bridge forms a parabola, of which the lowest part of the cable is 3 m above the surface of the bridge. The bridge is 90 M long and has a 8 m high Towers at the end of the bridge that supports the suspension cable at the top of each tower. if a quadratic function can model the cable, what would be coffient of x^2

Answer 1

Answer:

The coefficient of x^2 is;

$a=\frac{1}{405}$

Explanation:

Given that the lowest point of the cable is 3m above the surface of the bridge.

The bridge is 90 m long and has a 8m high towerat each edge.

Let us represent the question on a drawing;

From the drawing above, we assumed that the coordinates of the lowest point is (0,0);

$(h,k)=(0,0)$

Recall that the formula for vertex equation is;

$y=a(x-h)^2+k$

substituting the value of h and k;

$\begin{gathered} y=a(x-0)^2+0 \\ y=ax^2\text{ -------1} \end{gathered}$

From the derived equation 1, we can find the value of a which is the coefficient of x^2 by substituting the corresponding values of x and y at a particular point on the graph.

At the end of the bridge, the coordinate of the top right edge is;

$(x,y)=(45,5)$

substituting into the equation;

$\begin{gathered} y=ax^2 \\ a=\frac{y}{x^2} \\ a=\frac{5}{45^2} \\ a=\frac{1}{405} \end{gathered}$

Therefore, the coefficient of x^2 is;

$a=\frac{1}{405}$