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.) y = x2 + 6x − ___
if the vertex is (-3,-25) you can write \[-25=(-3)^2+6\times (3)+c\] and solve for c
or you could write \[0=(-8)^2+6\times (-8)+c\] and solve for c
for that matter you can write \[y=(x+8)(x-2)\] multiply out and find c. that may be easiest. \[y=x^2+6x-16\]
in fact the very very simplest way to do it is take the zeros, -8 and 2, and multiply them together to get \[c=-16\]
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