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x=-10 and x=-8 and a vertex at (-9, -3). Leave your answer in the form: f(x)=(a)(x^2+bx+c)
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i am going to assume that the first two are the zeros start with \[f(x)=a(x+10)(x+8)\] and then find \(a\) using the vertex
alternatively you could write \(f(x)=a(x+9)^2-3\) and solve for \(a\) using either zero
lets do the first method \[f(x)=a(x+10)(x+8)\] \[f(-9)=-3=a(-9+10)(-9+8)=a\times 1\times -1\] so we know \[-a=-3\] and therefore \(a=3\) and you get \[f(x)=3(x+10)(x+8)\]
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