6)
A quadratic function has the form
y = ax^2 + bx + c
Use point (3, 5) in the equation above:
5 = a(3^2) + 3b + c
5 = 9a + 3b + c
9a + 3b + c = 5 Equation 1
Use point (4, 3) in the equation above:
3 = a(4^2) + 4b + c
16a + 4b + c = 3 Equation 2
Use point (5, 3) in the equation above.
5 = a(5^2) + 5b + c
25a + 5b + c = 5 Equation 3.
Now solve the system of equations of equations 1, 2, and 3 to find the coefficients, a, b, and c.
9a + 3b + c = 5
16a + 4b + c = 3
25a + 5b + c = 5
Subtract the first equation from the second equation.
Subtract the second equation from the third equation.
You get
7a + b = -2
9a + b = 2
Subtract the first equation above from the second equation to get.
2a = 4
a = 2
Substitute:
7a + b = -2
7(2) + b = -2
b = -16
9a + 3b + c = 5
9(2) + 3(-16) + c = 5
18 - 48 + c = 5
c - 30 = 5
c = 35
The equation in standard form is
y = 2x^2 - 16x + 35
We can find it in vertex form:
y = 2(x^2 - 8x) + 35
y = 2(x^2 - 8x + 16) + 35 - 32
y = 2(x - 4)^2 + 3