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F(x)=x^2 + 4x + 12 what is the vertex form of f(x)? What is the minimum value of f(x)?

Sagot :

the equation of f(x) which is "x" terms is the equation of a vertical parabola that is opening upwards like a bowl, the minimum will be at its vertex

[tex]\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{1}x^2\stackrel{\stackrel{b}{\downarrow }}{+4}x\stackrel{\stackrel{c}{\downarrow }}{+12} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 4}{2(1)}~~~~ ,~~~~ 12-\cfrac{ (4)^2}{4(1)}\right) \implies \left( - \cfrac{ 4 }{ 2 }~~,~~12 - \cfrac{ 16 }{ 4 } \right) \\\\\\ (-2~~,~~12-4)\implies \stackrel{minimum}{(-2~~,~~\stackrel{\downarrow }{8})}[/tex]

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