Question 358553
Since ({{{1}}},{{{3}}}) and ({{{2}}},{{{3}}}) share a y-value, they are on opposite sides of the vertex.
The {{{x}}} coordinate of the vertex is {{{(1+2)/2=3/2}}}
Use the vertex form,
{{{y=a(x-h)^2+k}}}
{{{y=a(x-3/2)^2+k}}}
Use your points to solve for {{{a}}} and {{{k}}}.
{{{3=a(1-3/2)^2+k}}}
{{{3=a(1/4)+k}}}
1.{{{a+4k=12}}}
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{{{5=a(3-3/2)^2+k}}}
{{{5=a(9/4)+k}}}
2.{{{9a+4k=20}}}
Subtract eq. 1 from eq. 2,
{{{9a+4k-a-4k=20-12}}}
{{{8a=8}}}
{{{highlight_green(a=1)}}}
Then use either equation to solve for {{{k}}}.
{{{1+4k=12}}}
{{{4k=11}}}
{{{highlight_green(k=11/4)}}}
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{{{y=a(x-3/2)^2+k}}}
{{{y=(x-3/2)^2+11/4}}}
{{{y=(x^2-3x+9/4)+11/4}}}
{{{highlight(y=x^2-3x+5)}}}
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{{{drawing(300,300,-2,8,-2,8,circle(3/2,11/4,0.2),circle(1,3,0.2),circle(2,3,0.2),circle(3,5,0.2),grid(1),blue(line(3/2,-10,3/2,10)),graph(300,300,-2,8,-2,8,x^2-3x+5))}}}