In the given graph, the coordinates of the vertex of the parabola is,
(h, k)=(6, 64)
The given parabola is open downwards.
The vertex form of a parabola openingupwards or downwards is given by,
[tex]y=a(x-h)^2+k\ldots\ldots(1)[/tex]
Here, a is a constant and (h, k) are the coordinates of the vertex of the parabola.
If a>0, parabola opens upwards and if a<0, theparabola opens downwards.
(x,y)=(14,0) is a point on the given parabola. So, put x=14, y=0 h=6 and k=64 in the above equation to find the value of a.
[tex]\begin{gathered} 0=a(14-6)^2+64 \\ 0=a\times8^2+64 \\ 0=a\times64+64 \\ \frac{-64}{64}=a \\ -1=a \end{gathered}[/tex]
Now, put the values of a, h and k in equation (1).
[tex]\begin{gathered} y=-1\times(x-6)^2+64 \\ =-(x-6)^2+64 \end{gathered}[/tex]
Now, rewriting the above equation,
[tex]\begin{gathered} y=-(x^2-2\times6x+6^2)+64^{} \\ =-(x^2-12x+36)+64 \\ =-x^2+12x-36+64 \\ =-x^2+12x+28 \\ =-(x^2-12x-28) \\ =-(x^2-14x+2x-14\times2) \\ =-(x(x-14)+2(x-14)) \\ =-(x+2)(x-14) \end{gathered}[/tex]
Therefore, the equation of parabola is,
[tex]y=-(x+2)(x-14)[/tex]
