91影视

The given diagram is shown below:

Assume the lower left side of the bridge be the origin of the coordinate grid at the point $(0,0)$.

From the given diagram, observe that the base is $60$feet wide which represents the lower right side at the point $(60,0)$.

Also, the bridge is $90$feet high at the highest point which can be represented as the vertex of the parabola. So, the $y$-coordinate of the vertex will be localid="1645445735745" $90$.

The vertex must fall halfway between the left most point, localid="1645445741791" $(0,0)$, and the rightmost point localid="1645445969513" $(60,0)$since the bridge is symmetric.

Thus, conclude that the $x$-coordinate of the vertex will be localid="1645445754641" $30$.

The graph is shown below:

$\begin{array}{rcl}(h,k)& =& (30,90)\\ & 鈬�& h=30\\ & 鈬�& k=90\end{array}$

The standard form of parabola is given by $y=a{(x-h)}^2+k鈥�\left(1\right)$.

Substitute the values of $h$and $k$into $\left(1\right)$.

role="math" localid="1645445095224" $y={a(x-30)}^2+90鈥�\left(2\right)$

Consider the point $(x,y)=(0,0)$.

Substitute the point $(0,0)$into the equation $\left(2\right)$.

role="math" localid="1645445337468" $\begin{array}{rcl}0& =& a{(0-30)}^2+90\\ -90& =& a{(-30)}^2\\ -90& =& 900a\\ \frac{-90}{900}& =& a\end{array}$

$鈬�\begin{array}{rcl}& & a\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & \frac{1}{10}\end{array}$

$y=-\frac{1}{10}{(x-30)}^2+90$.