Question 217107
a  rough  sketch  helps,,,,,
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On  x-y  coordinates, line  if  symmetry  is  x=4,  so  draw  a  vertical  at  +4
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We  also  know  that  the  vertex  is  on  the  line  of  symmetry  so (4,?) is  vertex
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But  they  give  y=6 at  vertex,,,,,therefore  vertex  is (4,6),,,might  draw  point
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Since  x  intercept  is  2,,,(1) we  know(2,0)  is  point  on  curve,
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(2) drawing  point  on  rough  graph,  shows  it  to  be  curve  pointing  down.
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The  basic  eqn  is  probably  a  y  parabola  pointing  down  ,,,,,,from  sketch
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base  eqn  is  y=x^2 +b
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expanded  form  is (y-k) = A(x-h)^2
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we  know  vertex is (4,6) ,,,therefore h=4,,,k=6
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subst,,,,(y-6)=A(x-4)^2,,,,,subst (2,0)
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{(0)-6} =A {(2)-4}^2
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-6 =A(-2)^2
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-6 =A(4)
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A=-6/4 = -3/2=-1.5,,,,,,note  negative  confirms  pointing  down parabola
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combining,,,,(y-6)=-1.5(x-4)^2,,,,,answer
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checking,,,,(2,0),,,,(0-6) =-1.5(2-4)^2,,,-6=-1.5(-2)^2 =-1.5*4=-6,,,ok
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and (4,6),,,,,,6-6= -1.5(4-4)^2,,,,or  0=0,,,,,ok
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in  standard  form,,,(y-6) = -1.5(x-4)^2
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y-6 =-1,5{x^2-8x+16) = -1.5x^2 +12x -24
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y= -1.5x^2 +12x -18,,,,,std  form  answer
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checking,,,vertex  x = -b/2a = -12/(2*-1.5) = +4 ,,,,,ok
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y=-1.5(4)^2 +12(4) -18 = -24 +48-18=6,,,,ok
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