Question 1148201
The general form is {{{ f(x) =  a*x^2 + b*x + c }}}
The y-component of ( 6, -1 ) is greater than the
y-component of the vertex, so I know the vertex is 
a minimum ( just an observation )
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( 4, -3 )
{{{ -3 = a*4^2 + b*4 + c }}}
(1) {{{ -3 = 16a + 4b + c }}}
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For the vertex:
{{{ -b/(2a) = 4 }}}
(2) {{{ b = -8a }}}
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( 6, -1 )
{{{ -1 = a*6^2 + b*6 + c }}}
(3) {{{ -1 = 36a + 6b + c }}}
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You have 3 equation with 3 unknowns, so
it's solvable
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Plug (2) into (1) and (3)
(1) {{{ 16a + 4*( -8a ) + c + 3 = 0 }}}
(1) {{{ 16a - 32a + c + 3 = 0 }}}
(1) {{{ -16a + c + 3 = 0 }}}
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(3) {{{ 36a + 6*( -8a ) + c + 1 = 0 }}}
(3) {{{ -12a + c + 1 = 0 }}}
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Subtract (1) from (3)
(3) {{{ -12a + c + 1 = 0 }}}
(1) {{{ 16a - c - 3 = 0 }}}
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{{{ 4a - 2 = 0 }}}
{{{ 4a = 2 }}}
{{{ a = 1/2 }}}
and
(2) {{{ b = -8a }}}
(2) {{{ b = -4 }}}
and
(3) {{{ -12a + c + 1 = 0 }}}
(3) {{{ -12*( 1/2) + c + 1 = 0 }}}
(3) {{{ -6 + c + 1 = 0 }}}
(3) {{{ c = 5 }}}
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The equation is:
{{{ y = (1/2)*x^2 - 4x + 5 }}}
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check:
( 4, -3 )
{{{ -3 = (1/2)*4^2 - 4*4 + 5 }}}
{{{ -3 = 8 - 16 + 5 }}}
{{{ -3 = -3 }}}
OK
and
( 6, -1 )
{{{ -1 = (1/2)*6^2 -4*6 + 5 }}}
{{{ -1 = 18 - 24 + 5 }}}
{{{ -1 = -1 }}}
OK
Here's the plot:
{{{ graph( 400, 400, -10, 10, -10, 10, (1/2)*x^2 - 4x + 5 ) }}}
Looks OK