Answer:
D. \(-x^{2} -2x+3=2x^{2} -8x+3\)
Step-by-step explanation:
We are given that,
The graph of the system of equations is a 'quadratic graph opening down and a quadratic graph opening up'.
This means that one quadratic equation will have leading co-efficient positive and other will have leading co-efficient negative.
So, we get that options A and B are discarded.
Further it is provided that the graph intersect at ( 0,3 ) and ( 2,-5 ).
This means that the pair of points must satisfy the given system of equations.
So, according to the options:
C. \(-x^{2} -2x+3=2x^{2} -8x-3\)
Putting x = 0, gives 3 = -3, which is not possible.
So, option C is dicarded.
D. \(-x^{2} -2x+3=2x^{2} -8x+3\)
Putting x = 0 gives 3 = 3 and x = 2 gives -5 = -5.
Hence, the graph of the given system solves the equation \(-x^{2} -2x+3=2x^{2} -8x+3\).
