Question 1032404
Solving in this case means finding
the roots, or all {{{ x }}} when {{{ f(x) = 0 }}}
{{{ 0 = -(x-3)^2 }}}
I can see that {{{ x=3 }}} solves this
{{{ 0 = -( 3-3 )^2 }}}
{{{ 0 = 0 }}}
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This is called a "double root" because a 
quadratic should have 2 solutions
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The fact that there is one solution is shown by 
both the graph and the quadratic formula:
{{{ x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
When there is just one solution, the discriminant is zero
{{{ b^2 - 4a*c = 0 }}}
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{{{ f(x) = -(x-3)^2 }}}
{{{ f(x) = -( x^2 - 6x + 9 ) }}}
{{{ f(x) = -x^2 + 6x - 9 }}}
{{{ a = 1 }}}
{{{ b = 6 }}}
{{{ c = -9 }}}
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The discriminant is:
{{{ b^2 - 4a*c = 6^2 - 4*(-1)*(-9) }}}
{{{ b^2 - 4a*c = 36 - 36 }}}
{{{ b^2 - 4a*c = 0 }}} 
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So, referring to quadratic formula, the
single root is at:
{{{ x = -b/(2a) }}}
{{{ x = -6/(2*(-1)) }}}
{{{ x = 3 }}} ( what we decided already )
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Here's the plot:
{{{ graph( 400, 400, -2, 10, -10, 2, -x^2 + 6x - 9 ) }}}