To find the polynomial function, we use the general form of a polynomial of degree 2:
y = ax^2 + bx + c
We can substitute the x and y values from each of the given points to get a system of equations:
(5, 13) -> 13 = 25a + 5b + c
(10, -11) -> -11 = 100a + 10b + c
(0, 3) -> 3 = 0a + 0b + c
The third equation implies that c = 3.
Now we can substitute the value of c in the first two equations:
13 = 25a + 5b + 3 -> 10 = 25a + 5b
-11 = 100a + 10b + 3 -> -14 = 100a + 10b
Simplifying these equations gives us the system:
10 = 25a + 5b
-14 = 100a + 10b
We can solve this system using elimination or substitution. Let's use elimination:
Multiply the first equation by 2 and the second equation by 5:
20 = 50a + 10b
-70 = 500a + 50b
Now subtract the second equation from the first:
90 = 450a
a = 90/450
a = 0.2
Substitute the value of a in one of the equations to solve for b:
10 = 25(0.2) + 5b
10 = 5 + 5b
5 = 5b
b = 1
Now we have the values of a, b, and c:
a = 0.2
b = 1
c = 3
Therefore, the polynomial function that passes through (5,13), (10,-11), and (0,3) is:
y = 0.2x^2 + x + 3