Question 371918
You can convert to vertex form.
{{{y=-(x-2)^2+1}}}
For y-intercept, set {{{x=0}}}, solve for {{{y}}}.
{{{y=-(0-2)^2+1}}}
{{{y=-4+1}}}
{{{y=-3}}}
({{{0}}},{{{-3}}})
For x-intercept, set {{{y=0}}}, solve for {{{x}}}.
{{{0=-(x-2)^2+1}}}
{{{(x-2)^2=1}}}
{{{x-2=0 +- 1}}}
{{{x=2 +- 1}}}
{{{x=3}}} and {{{x=1}}}
({{{3}}},{{{0}}}) and ({{{1}}},{{{0}}})
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{{{drawing(300,300,-2,6,-4,4,grid(1),
circle(2,1,0.2),
circle(3,0,0.2),
circle(1,0,0.2),
circle(0,-3,0.2),
graph(300,300,-2,6,-4,4,0,-(x-2)^2+1,-x^2+4x-3))}}}