Question 524747
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Since the vertex is on the *[tex \Large y]-axis, symmetry says that if *[tex \Large (-1, 6)] is on the parabola, *[tex \Large (1,6)] is on the parabola.  Now that we have three points on the parabola we can write the equation.


All parabolas with a vertical axis of symmetry can be described by a function of the form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(x)\ =\ ax^2\ +\ bx\ +\ c]


So, if the point *[tex \Large (0,1)] is on the parabola, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(0)^2\ +\ b(0)\ +\ c\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c\ =\ 1]


Then, if the point *[tex \Large (-1,6)] is on the parabola, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(-1)^2\ +\ b(-1)\ +\ c\ =\ 6]


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


Then, if the point *[tex \Large (1,6)] is on the parabola, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(1)^2\ +\ b(1)\ +\ c\ =\ 6]


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


Substituting the known value of *[tex \Large c] we are left with a 2X2 system:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ -\ b\ =\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ +\ b\ =\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2a\ =\ 10]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ =\ 0]


And then substituting the now known coefficients into the function definition:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y=\ \rho(x)\ =\ 5x^2\ +\ 1]


Since the lead coefficient is positive, the parabola opens upward.


{{{drawing(
500, 500, -3,3,-1,19,
grid(1),
graph(
500, 500, -3,3,-1,19,
5x^2+1))}}}


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