Question 725980
If equation has the form {{{ y = a*x^2 + b*x + c }}},
then the x co-ordinate of the vertex is at
{{{ x[v] = -b/(a) }}}
given: vertex at (2,-4)
{{{ -b/(2a) = 2 }}}
{{{ -b = 4a }}}
{{{ b = -4a }}}
also:
{{{ -4 = a*x^2 - 4a*x + c }}}
{{{ -4 = a*2^2 - 4a*2 + c }}}
{{{ -4 = 4a - 8a + c
{{{ -4a + c = -4 }}}
(1) {{{ 4a - c = 4 }}}
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Also given: (-1,3) is a solution
{{{ y = a*x^2 + b*x + c }}}
{{{ 3 = a*(-1)^2 - 4a*(-1) + c }}}
{{{ 3 = a + 4a + c }}}
(2) {{{ 5a + c = 3 }}}
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Add (1) and (2)
(1) {{{ 4a - c = 4 }}}
(2) {{{ 5a + c = 3 }}}
{{{ 9a = 7 }}}
{{{ a = 7/9 }}}
and, since
{{{ b = -4a }}}
{{{ b = -4*(7/9) }}}
{{{ b = -28/9 }}}
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also
(2) {{{ 5a + c = 3 }}}
(2) {{{ 35/9 + c = 3 }}}
(2) {{{ c = 27/9 - 35/9 }}}
(2) {{{ c = -8/9 }}}
The equation is:
{{{ y = (7/9)*x^2 - (28/9)*x - 8/9 }}}
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check: is the vertex at (2,-4) ?
{{{ -b/(2a) = 2 }}}
{{{ (28/9) / (2*(7/9)) = 2 }}}
{{{ (28/9)*(9/14) = 2 }}}
{{{ 2 = 2 }}}
{{{ -4 = (7/9)*4 - (28/9)*2 - 8/9 }}}
{{{ -4 = 28/9 - 56/9 - 8/9 }}}
{{{ -36 = 28 -56 -8 }}}
{{{ -36 = -36 }}}
OK
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Does it go through (-1,3) ?
{{{ y = (7/9)*x^2 - (28/9)*x - 8/9 }}}
{{{ 3 = (7/9)*(-1)^2 - (28/9)*(-1) - 8/9 }}} 
{{{ 3 = 7/9 + 28/9 - 8/9 }}}
{{{ 27 = 7 + 28 - 8 }}}
{{{ 27 = 27 }}}
OK