Answer:
[tex]y=-\dfrac{13}{49}x^2[/tex]
Step-by-step explanation:
The shape of an arch corresponds to a parabola.
the general equation for a parabola is:
[tex]y=ax^2+bx+c[/tex]
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)
[tex]-13=a(-7)^2+b(-7)+c[/tex]
[tex]49a-7b+c=-13[/tex]
(x,y) = (7,-13)
[tex]-13=a(7)^2+b(7)+c[/tex]
[tex]49a+7b+c=-13[/tex]
(x,y) = (0,0)
[tex]0=a(0)^2+b(7)+c[/tex]
[tex]c=0[/tex]
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.
[tex]49a-7b+c=-13\quad\Rightarrow\quad49a-7b=-13[/tex]
[tex]49a+7b+c=-13\quad\Rightarrow\quad49a+7b=-13[/tex]
we can solve these two equation using the elimination method, by simply adding the two equations
[tex]\quad\quad49a-7b=-13\\+\quad49a+7b=-13[/tex]
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[tex]\quad\quad 98a=-26[/tex]
[tex]\quad\quad a=-\dfrac{13}{49}[/tex]
Now we can plug this value of a in any of the two equations.
[tex]49a-7b=-13[/tex]
[tex]49\left(-\dfrac{13}{49}\right)-7b=-13[/tex]
[tex]-13-7b=-13[/tex]
[tex]-7b=0[/tex]
[tex]b=0[/tex]
We have the values of a,b, and c. We can plug them in the general equation to find the equation of the arch.
[tex]y=\left(-\dfrac{13}{49}\right)x^2+0x+0[/tex]
[tex]y=-\dfrac{13}{49}x^2[/tex]
[tex]49y=-13x^2[/tex]
This our equation of the arch!
