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 − ___
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you know it looks like
\[a(x+8)(x-2)=a(x^2+6x-16)\] so you know the constant is -16
i said TMI because the first coordinate is an unnecessary piece of information
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ohkay thanks and are you sure that is the correct answer ?
you are asked to enter it without the minus sign, so put in "16"
the point is if the vertex was not (-3,-25) there would be no solution to this, so it is a rather dumb question. you are given more information that the poor quadratic can handle. you might want to mention this to your math teacher
since the vertex is (-3,-25) you know it looks like
\[y=a(x+3)^2-25\] and now all you need is one more point to find out that a = 1
in other words: the vertex and one other point would give the answer!