Question 381581
{{{f(x)=-x^2+4x-1}}}
{{{f(x)=-(x^2-4x+4)-1+4}}}
{{{f(x)=-(x-2)^2+3}}}
({{{2}}},{{{3}}}) is correct for the vertex.
The vertex lies on the axis of symmetry, {{{x=2}}}.
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To find the x-intercepts, now set {{{f(x)=0}}}.
{{{-(x-2)^2+3=0}}}
{{{(x-2)^2=3}}}
{{{x-2=0 +- sqrt(3)}}}
{{{highlight(x=2+- sqrt(3))}}}
Those are the x-intercepts.
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To find the y-intercepts, set {{{x=0}}} solve for {{{y}}}.
{{{f(x)=-(0-2)^2+3}}}
{{{f(x)=-4+3}}}
{{{f(x)=-1}}}
({{{0}}},{{{-1}}})
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There are no restrictions on the domain.
Since the parabola opens downwards, the value at the vertex is a maximum.
Domain:({{{-infinity}}},{{{infinity}}})
Range:({{{-infinity}}},{{{3}}}]

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{{{drawing(300,300,-6,6,-6,6,grid(1),blue(line(2,-100,2,100)),circle(0,-1,0.3),circle(2,3,0.3),circle(2+sqrt(3),0,0.3),circle(2-sqrt(3),0,0.3),graph(300,300,-6,6,-6,6,0,-(x-2)^2+3))}}}