Question 878392
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Why do you think it is b?


If *[tex \Large y\ =\ -x^2\ +\ 5]

Then


*[tex \Large y(-1)\ =\ -(-1)^2\ +\ 5\ =\ -1\ +\ 5\ =\ 4]


*[tex \Large y(0)\ =\ -(0)^2\ +\ 5\ =\ 0\ +\ 5\ =\ 5]


*[tex \Large y(1)\ =\ -(1)^2\ +\ 5\ =\ -1\ +\ 5\ =\ 4]


*[tex \Large y(2)\ =\ -(2)^2\ +\ 5\ =\ -4\ +\ 5\ =\ 1]


*[tex \Large y(3)\ =\ -(3)^2\ +\ 5\ =\ -9\ +\ 5\ =\ -4]


So apparently b is incorrect.


Consider the standard form of a quadratic function:


*[tex \Large y\ =\ ax^2\ +\ bx\ +\ c]


If the point *[tex \Large (-1, 6)] is an element of the solution set, then


*[tex \Large 6\ =\ a(-1)^2\ +\ b(-1)\ +\ c] must hold and


If the point *[tex \Large (0, 5)] is an element of the solution set, then


*[tex \Large 5\ =\ a(0)^2\ +\ b(0)\ +\ c] must hold and


If the point *[tex \Large (1, 6)] is an element of the solution set, then


*[tex \Large 6\ =\ a(1)^2\ + b(1)\ +\ c] must hold.


Simplfying:


*[tex \Large a\ -\ b\ +\ c\ =\ 6]


*[tex \Large c\ =\ 5]


*[tex \Large a\ +\ b\ +\ c\ =\ 6]


Substituting


*[tex \Large a\ -\ b\ =\ 1]


*[tex \Large a\ +\ b\ =\ 1]


Solving by elimination:


*[tex \Large a\ =\ 1] and *[tex \Large b\ =\ 0]


Substituting the now known values of a, b, and c into *[tex \Large ax^2\ +\ bx\ +\ c] yields:


*[tex \Large y\ =\ x^2\ +\ 5]


Checking the result:


*[tex \Large y(-1)\ =\ (-1)^2\ +\ 5\ =\ 1\ +\ 5\ =\ 6]


*[tex \Large y(0)\ =\ (0)^2\ +\ 5\ =\ 0\ +\ 5\ =\ 5]


*[tex \Large y(1)\ =\ (1)^2\ +\ 5\ =\ 1\ +\ 5\ =\ 6]


*[tex \Large y(2)\ =\ (2)^2\ +\ 5\ =\ 4\ +\ 5\ =\ 9]


*[tex \Large y(3)\ =\ (3)^2\ +\ 5\ =\ 9\ +\ 5\ =\ 14]


Looks right to me.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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