Question

Solve for vertex algebraically x^2+4x+6

Answer 1

`\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 }}{+6} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 4}{2(1)}~~~~ ,~~~~ 6-\cfrac{ (4)^2}{4(1)}\right) \implies \left( - \cfrac{ 4 }{ 2 }~~,~~6 - \cfrac{ 16 }{ 4 } \right) \\\\\\ \left( -2 ~~~~ ,~~~~ 6 -4 \right)\implies (-2~~,~~2)`