Question

Find the axis of symmetry for f(x)=x^{2}+12x+31

Answer 1

so since the squared variable is the "x", we're looking at a vertical parabola, and its axis of symmetry will be at the x-coordinate for its vertex, let's find its vertex then

`\textit{vertex of a vertical parabola, using coefficients} \\\\ f(x)=\stackrel{\stackrel{a}{\downarrow }}{1}x^2\stackrel{\stackrel{b}{\downarrow }}{+12}x\stackrel{\stackrel{c}{\downarrow }}{+31} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)`

`\left(-\cfrac{ 12}{2(1)}~~~~ ,~~~~ 31-\cfrac{ (12)^2}{4(1)}\right) \implies \left( - \cfrac{ 12 }{ 2 }~~,~~31 - \cfrac{ 144 }{ 4 } \right) \\\\\\ \left( -6 ~~~~ ,~~~~ 31 -36 \right)\implies (\stackrel{ x }{-6}~~,~-5)\hspace{5em}\stackrel{ \textit{axis of symmetry} }{x=-6}`