Question

Select the correct answer.What is the equation of the parabola shown with its focus on this graph?

Answer 1

Given:

From the graph, the vertex is at (0,1)

Using the equation of a parabola in vertex form;

`\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}`

To get the constant a,

`\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 \\ -(4)/(49)=a \\ a\approx-(1)/(12) \end{gathered}`

Hence, the equation is;

`y=-(1)/(12)x^2+1`

The graph is shown;

Therefore, option A is correct.