Question

Write the equation of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

Short answer

The equation in standard form is $y=-\frac{1}{5}{(x-15)}^2+45$.

Step-by-step solution

Step 1. Given information

The given diagram is shown below:

Step 2. Set up a coordinate system from the given diagram.

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 $30$feet wide which represents the lower right side at the point $(30,0)$.

Also, the bridge is $45$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 $45$.

The vertex must fall halfway between the left most point, $(0,0)$, and the rightmost point $(30,0)$since the bridge is symmetric.

Thus, conclude that the $x$-coordinate of the vertex will be $15$.

The graph is shown below:

Step 3. Determine the vertex, (h,k). 

$\begin{array}{rcl}(h,k)& =& (15,45)\\ & \Rightarrow & h=15\\ & \Rightarrow & k=45\end{array}$

Substitute the values of h and k into the standard form of parabola.

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

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

role="math" localid="1645447205539" $y=a{(x-15)}^2+45\dots \left(2\right)$

Find the value of a by using one of the other points on the parabola. 

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

Substitute the point $(0,0)$into the equation localid="1645447676800" $\left(2\right)$.

localid="1645447439894" $\begin{array}{rcl}0& =& a{(0-15)}^2+45\\ -45& =& a{(-15)}^2\\ -45& =& 225a\\ \frac{-45}{225}& =& a\end{array}$

$\Rightarrow a=-\frac{1}{5}$

Substitute the value for a into the equation (2).

$y=-\frac{1}{5}{(x-15)}^2+45$