Question

Find the vertex of the function given below.

y = 3x² + 6x +1
A. (1,7)
B. (-1,-2)
C. (-4,9)
D. (-1,-1)

Answer 1

`\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{3}x^2\stackrel{\stackrel{b}{\downarrow }}{+6}x\stackrel{\stackrel{c}{\downarrow }}{+1} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 6}{2(3)}~~~~ ,~~~~ 1-\cfrac{ (6)^2}{4(3)}\right) \implies \left( - \cfrac{ 6 }{ 6 }~~,~~1 - \cfrac{ 36 }{ 12 } \right) \\\\\\ (-1~~,~~1-3)\implies {\Large \begin{array}{llll} (-1~~,~~-2) \end{array}}`