Answer:
![y=-\dfrac{13}{49}x^2](https://tex.z-dn.net/?f=y%3D-%5Cdfrac%7B13%7D%7B49%7Dx%5E2)
Step-by-step explanation:
The shape of an arch corresponds to a parabola.
the general equation for a parabola is:
![y=ax^2+bx+c](https://tex.z-dn.net/?f=y%3Dax%5E2%2Bbx%2Bc)
we're given three coordinates: (-7,-13),(7,-13) and (0,0)
so we can plug these values in the general equation to make 3 separate equations:
(x,y) = (-7,-13)
![-13=a(-7)^2+b(-7)+c](https://tex.z-dn.net/?f=-13%3Da%28-7%29%5E2%2Bb%28-7%29%2Bc)
![49a-7b+c=-13](https://tex.z-dn.net/?f=49a-7b%2Bc%3D-13)
(x,y) = (7,-13)
![-13=a(7)^2+b(7)+c](https://tex.z-dn.net/?f=-13%3Da%287%29%5E2%2Bb%287%29%2Bc)
![49a+7b+c=-13](https://tex.z-dn.net/?f=49a%2B7b%2Bc%3D-13)
(x,y) = (0,0)
![0=a(0)^2+b(7)+c](https://tex.z-dn.net/?f=0%3Da%280%29%5E2%2Bb%287%29%2Bc)
![c=0](https://tex.z-dn.net/?f=c%3D0)
so we have three equations. and we can solve them simultaneously to find the values of a,b, and c.
we've already found c = 0, let's use substitute it to other equations.
![49a-7b+c=-13\quad\Rightarrow\quad49a-7b=-13](https://tex.z-dn.net/?f=49a-7b%2Bc%3D-13%5Cquad%5CRightarrow%5Cquad49a-7b%3D-13)
![49a+7b+c=-13\quad\Rightarrow\quad49a+7b=-13](https://tex.z-dn.net/?f=49a%2B7b%2Bc%3D-13%5Cquad%5CRightarrow%5Cquad49a%2B7b%3D-13)
we can solve these two equation using the elimination method, by simply adding the two equations
![\quad\quad49a-7b=-13\\+\quad49a+7b=-13](https://tex.z-dn.net/?f=%5Cquad%5Cquad49a-7b%3D-13%5C%5C%2B%5Cquad49a%2B7b%3D-13)
------------------------------
![\quad\quad 98a=-26](https://tex.z-dn.net/?f=%5Cquad%5Cquad%2098a%3D-26)
![\quad\quad a=-\dfrac{13}{49}](https://tex.z-dn.net/?f=%5Cquad%5Cquad%20a%3D-%5Cdfrac%7B13%7D%7B49%7D)
Now we can plug this value of a in any of the two equations.
![49a-7b=-13](https://tex.z-dn.net/?f=49a-7b%3D-13)
![49\left(-\dfrac{13}{49}\right)-7b=-13](https://tex.z-dn.net/?f=49%5Cleft%28-%5Cdfrac%7B13%7D%7B49%7D%5Cright%29-7b%3D-13)
![-13-7b=-13](https://tex.z-dn.net/?f=-13-7b%3D-13)
![-7b=0](https://tex.z-dn.net/?f=-7b%3D0)
![b=0](https://tex.z-dn.net/?f=b%3D0)
We have the values of a,b, and c. We can plug them in the general equation to find the equation of the arch.
![y=\left(-\dfrac{13}{49}\right)x^2+0x+0](https://tex.z-dn.net/?f=y%3D%5Cleft%28-%5Cdfrac%7B13%7D%7B49%7D%5Cright%29x%5E2%2B0x%2B0)
![y=-\dfrac{13}{49}x^2](https://tex.z-dn.net/?f=y%3D-%5Cdfrac%7B13%7D%7B49%7Dx%5E2)
![49y=-13x^2](https://tex.z-dn.net/?f=49y%3D-13x%5E2)
This our equation of the arch!