The Standard Form of a Quadratic Function:

Using the given points (-1,5), (0,3), and (3,9), let's substitute each point to the equation.
At (-1,5):


At (0,3):


At (3,9):


We now get these equations:
5 = a - b + c ; 3 =c; 9 = 9a + 3b + c
Let's determine the value of a, b and c. We get,
Substituting 3 = c to 5 = a - b + c,


Let's substitute 3 = c and b = a - 2 to 9 = 9a + 3b + c,



Since a = 1, let's solve for the value of b which is b = a - 2.


Since we've identified that a = 1, b = -1 and c = 3, let's substitute the values to the standard form of a quadratic function to be able to make the equation.


Therefore, the quadratic function in a standard form whose graph passes through the given points (-1,5), (0,3), (3,9) is y = x^2 - x + 3.