You know, there are 2 types of standard form of the equation of parabola.

a) Equation form: \(y=ax^2+bx+c\) (When a>0, the parabola opens upward & When a<0, the parabola opens downward)
b) Vertex form: \(y=a(x-h)^2+k\), where (h,k) is the vertex.

Now, observe from the given figure that the parabola is in vertex form with (h,k)=(negative constant,positive constant) OR "The vertex(h,k) lies in Q2". Let's go through each of the options and compare the equations with the vertex form of parabolan to find (h,k). The equation with (h,k) in Q2 would be our final answer.

A. \(y = 1(x-0)^2 + 3\); here (h,k)=(0,3), which is on the axis. DISCARD
B. \(y = 1(x - 3)^2 + 3\); here (h,k)=(3,3), which is on the Q1. DISCARD
C. \(y = 1(x - (-3))^2 + (-3)\); here (h,k)=(-3,-3), which is on the Q3. DISCARD
D. \(y = 1(x - 3)^2 +(-3)\);here (h,k)=(3,-3), which is on the Q4. DISCARD
E. \(y = (x - (-3))^2 + 3\);here (h,k)=(-3,3), which is on the Q2. This is our desired equation of the given parabola.

Ans. (E)
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Regards,

PKN

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