Rewrite the following quadratic functions in form f(x) = a(x-h) ²+ k

[tex] \tt{1.)y = {x}^{2} + 8x + 12 } \\ [/tex]
[tex]\tt y + ( \frac{8}{2} ) {}^{2} = ( {x}^{2} + 8x + ( \frac{8}{2} ) {}^{2} ) + 12[/tex]
[tex]\tt y + 16 = ( {x}^{2} + 8x +16 ) + 12 \\ \tt y + 16 = ( {x} + 4) {}^{2} + 12 \\\tt y = ( {x} + 4) {}^{2} + 12 - 16 \\ \boxed{\red{\tt y = ( {x} + 4) {}^{2} - 4}}[/tex]
[tex] \red{﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌}[/tex]
[tex] \tt{}2.)y = - x {}^{2} - 2x - 70 [/tex]
[tex] \small\tt{y - ( \frac{ - 2}{2}) {}^{2} = -( x {}^{2} + 2x + ( \frac{ - 2}{2}) {}^{2} ) - 70}[/tex]
[tex] \tt{y - 1 = - ({x}^{2} + 2x + 1) } - 70 \\ \tt{y= - ({x} + 1) }^{2} - 70 + 1 \\ \boxed{\red{\tt{y= - ({x} + 1) }^{2} - 69}}[/tex]
[tex] \red{﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌}[/tex]
[tex] \tt3.)y = {x}^{2} - 5x - 6 \\ [/tex]
[tex] \tt{y + ( \frac{ - 5}{2} ) {}^{2} = ({x}^{2} - 5x + ( \frac{ - 5}{2} ) {}^{2})- 6} \\ [/tex]
[tex]y + \frac{25}{4} = ({x}^{2} - 5x + \frac{25}{4} )- 6 \\ y + \frac{25}{4} = ({x} - \frac{5}{2} ) {}^{2} - 6 \\ y = {(x - \frac{5}{2}) }^{2} - 6 - \frac{25}{4} \\ \boxed{ \red{ \tt{y = (x - \frac{5}{2} ) {}^{2} - \frac{49}{4} }}}[/tex]
[tex] \red{﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌}[/tex]
[tex] \tt4.)y = 3 {x}^{2} - 2x + 11 \\ [/tex]
[tex]y = 3( \frac{3 {x}^{2} - 2x }{3} ) + 11 \\ y = 3( {x}^{2} - \frac{2}{3} x) + 11 \\ [/tex]
[tex]y + 3( \frac{1}{9} ) = 3( {x}^{2} - \frac{2}{3} + \frac{1}{9} ) + 11 \\ y + \frac{1}{3} = 3(x - \frac{1}{3} ) {}^{2} + 11 \\ y = 3(x - \frac{1}{3}) {}^{2} + 11 - \frac{1}{3} \\ \boxed{ \red{ \tt{y = 3(x - \frac{1}{3}) {}^{2} + \frac{32}{3} }}}[/tex]
[tex] \red{﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌﹌}[/tex]