EXPLANATION
Given the function f(x) = (x-6)^2 + 1
[tex]\mathrm{The\: vertex\: of\: an\: up-down\: facing\: parabola\: of\: the\: form}\: y=ax^2+bx+c\: \mathrm{is}\: x_v=-\frac{b}{2a}[/tex]Expanding (x-6)^2 + 1 by applying the Perfect Square Formula:
[tex]=x^2-12x+37[/tex][tex]\mathrm{The\: parabola\: params\: are\colon}[/tex][tex]a=1,\: b=-12,\: c=37[/tex][tex]x_v=-\frac{b}{2a}[/tex][tex]x_v=-\frac{\left(-12\right)}{2\cdot\:1}[/tex][tex]\mathrm{Simplify}[/tex][tex]x_v=6[/tex][tex]y_v=6^2-12\cdot\: 6+37[/tex]Simplify:
[tex]y_v=1[/tex][tex]\mathrm{Therefore\: the\: parabola\: vertex\: is}[/tex][tex]\mleft(6,\: 1\mright)[/tex][tex]\mathrm{If}\: a<0,\: \mathrm{then\: the\: vertex\: is\: a\: maximum\: value}[/tex][tex]\mathrm{If}\: a>0,\: \mathrm{then\: the\: vertex\: is\: a\: minimum\: value}[/tex][tex]a=1[/tex][tex]\mathrm{Minimum}\mleft(6,\: 1\mright)[/tex][tex]\mathrm{For\: a\: parabola\: in\: standard\: form}\: y=ax^2+bx+c\: \mathrm{the\: axis\: of\: symmetry\: is\: the\: vertical\: line\: that\: goes\: through\: the\: vertex}\: x=\frac{-b}{2a}[/tex]Expanding (x-6)^2 + 1 by applying the Perfect Square Formula:
[tex]y=x^2-12x+37[/tex][tex]\mathrm{Axis\: of\: Symmetry\: for}\: y=ax^2+bx+c\: \mathrm{is}\: x=\frac{-b}{2a}[/tex][tex]a=1,\: b=-12[/tex][tex]x=\frac{-\left(-12\right)}{2\cdot\:1}[/tex][tex]\mathrm{Refine}[/tex]Axis of simmetry : x=6
The quadratic function has the same shape than the parent function y=x^2 because there is NOT a coefficient within x.
