Question

fill in the missing term so that the quadratic equation has a graph that opens up with a vertex of (- 3, -25), and x intercepts at x= -8 and x=2. ( do not include the negative sign in your answer.)

Answer 1

\[y=ax^2+bx+c\]
0=a(2)^2+b(2)+c
0=a(-8)^2+b(-8)+c
-25=a(-3)^2+b(-3)+c

Answer 2

Solve the system

Answer 3

\[4a^2+2b+c\]=0
\[64a-8b+c=0\]
\[9a-3b+c=-25\]