Question 556161
Two points determine a linear function; three points determine a quadratic function.
A quadratic function can be written as
{{{f(x)=ax^2+bx+c}}} or {{{y=ax^2+bx+c}}}
In general, you can always substitute the coordinates of each of 3 points to get 3 equations.
For your problem,
(-1,1) ---> {{{1=a(-1)^2+b(-1)+c}}} ---> {{{1=a-b+c}}}
(1,1) ---> {{{1=a(1)^2+b(1)+c}}} ---> {{{1=a+b+c}}}
(3,9) ---> {{{9=a(3)^2+b(3)+c}}} ---> {{{9=9a+3b+c}}}
You got a system of equations.
From there, you solve the system for a, b, and c and those coefficients determine your quadratic function.
In your case, symmetrical points (-1, 1), and (1, 1) tell you that the axis of symmetry will be the y-axis (the line x=0), making b=0.
In your case {{{f(x)=ax^2+c}}} or {{{y=ax^2+c}}}
The simplest quadratic function (the mother of all quadratic functions) is
{{{f(x)=x^2}}} or {{{y=x^2}}}
Without grabbing your pencil (or pen), you can see that it passes through all 3 of your points. There is only one quadratic function that passes through any set of 3 points, so
Your function is {{{f(x)=x^2}}} or {{{y=x^2}}}.
(But you can solve the system of equations if it makes you, or your teacher happy. It's an easy one.)