Question 982271
<font face="Times New Roman" size="+2">


With the vertex at (1,-13) and another point at (0,-11), by symmetry a third point must be at (2,-11).


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


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  a(0)^2\ +\ b(0)\ +\ c\ =\ -11]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  a(1)^2\ +\ b(1)\ +\ c\ =\ -13]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  a(2)^2\ +\ b(2)\ +\ c\ =\ -11]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  c\ =\ -11]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  a\ +\ b\ +\ c\ =\ -13]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  4a\ +\ 2b\ +\ c\ =\ -11]


Solve the 3X3 system of equations to get the coefficients a, b, and c.


Substitute the coefficients in


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  ax^2\ +\ bx\ + c\ =\ 0]


and solve the quadratic for the two zeros to get the *[tex \Large x]-coordinates of the *[tex \Large x]-intercepts.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \