Solution:
Given:
From the graph, the vertex is at (0,1)
Using the equation of a parabola in vertex form;
[tex]\begin{gathered} y=a(x-h)^2+k \\ where; \\ (h,k)=(0,1) \\ h=0 \\ k=1 \\ \\ Hence, \\ y=a(x-0)^2+1 \\ y=ax^2+1 \end{gathered}[/tex]
To get the constant a,
[tex]\begin{gathered} Using\text{ the point }(7,-3) \\ x=7 \\ y=-3 \\ \\ y=ax^2+1 \\ -3=a(5^2)+1 \\ -3-1=49a \\ -4=49a \\ -\frac{4}{49}=a \\ a\approx-\frac{1}{12} \end{gathered}[/tex]
Hence, the equation is;
[tex]y=-\frac{1}{12}x^2+1[/tex]
The graph is shown;
Therefore, option A is correct.
