Question 1002233


Find the equation in the form of {{{a(x-h)^2+k}}}: it is a parabola
Turning point: ({{{1}}},{{{1}}}) -> turning point of a parabola is vertex
so, ({{{1}}},{{{1}}})=({{{h}}},{{{k}}})

y-intercept: {{{0 }}}=>({{{0}}},{{{0}}}) 

{{{y=a(x-h)^2+k}}}....if ({{{1}}},{{{1}}})=({{{h}}},{{{k}}})

{{{y=a(x-1)^2+1}}}......eq.1

{{{y=a(x-1)^2+1}}}....if ({{{0}}},{{{0}}})

{{{0=a(0-1)^2+1}}}

{{{-1=a(1)}}}

{{{-1=a}}}


{{{y=-1(x-1)^2+1}}} or

{{{y=-(x-1)^2+1}}}


{{{ graph( 600, 600, -10, 10, -10, 10, -(x-1)^2+1) }}}