[tex]\begin{gathered} \text{Let the quadractic function be } \\ y=x^2+bx+c \end{gathered}[/tex]
Given that the zeros of the function are at x= 2 and x=6
[tex]\begin{gathered} \text{When x=2, we have} \\ 0=2^2+b(2)+c \\ 0=4+2b+c \\ 2b+c=-4--------\text{equation (1)} \end{gathered}[/tex][tex]\begin{gathered} \text{When x = 6, we have} \\ 0=6^2+b(6)+c \\ 0=36+6b+c \\ 6b+c=-36-------------\text{equation}(2) \end{gathered}[/tex]
Solving both equations by elimination method, we have
2b + c = -4
6b+c = -36
2b + c-(6b+c)= -4 - (-36)
2b + c-6b-c=-4+36
-4b = 32
b= 32/-4
b= -8
put b= -8 into equation (1)
2b + c = -4
2( -8) + c = -4
-16 + c = -4
c= -4 + 16
c = 12
The function is
[tex]f(x)=x^2-8x+12[/tex]
The first missing value is -8, while the second missing value is 12
