Question 671570
Find a quadratic function in standard form for each set of points.
(-1,6), (1,4), (2,9)


{{{y=ax^2+bx+c}}}

{{{6=a(-1)^2+b(-1)+c}}} ........ for (-1,6)

{{{6=a-b+c}}}..........1

{{{4=a+ b+c}}} ..........2............for (1,4)

{{{9=4a+2b+c}}} .............3.........for (2,9)
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{{{6=a-b+c}}}..........1

{{{4=a+ b+c}}} ............for (1,4)..........2...add both 1 and 2
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{{{10=2a+2c}}}..........2a.........both sides divide by {{{2}}} 

{{{5=a+c}}}......solve for {{{a}}}

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{{{a=5-c}}}.......substitute in 3

{{{9=4(5-c)+2b+c}}}

{{{9=20-4c+2b+c}}}

{{{9=20-3c+2b}}}.....go to 2 and solve for {{{b}}}

{{{4-a-c=b}}} ..........2.......substitute  a=5-c

{{{4-(5-c)-c=b}}}

{{{4-5+c-c=b}}}

{{{highlight(-1=b)}}}

now find {{{c}}}

{{{9=20-3c+2(-1)}}}

{{{9=20-3c-2}}}

{{{9=18-3c}}}

{{{3c=18-9}}}

{{{3c=9}}}

{{{highlight(c=3)}}}

now find {{{a}}}

{{{a=5-c}}}

{{{a=5-3}}}

{{{highlight(a=2)}}}

so, your equation is: {{{y=2x^2-x+3}}}


{{{drawing(600,600,-10,10,-10,10,grid(1),circle(-1,6,0.2),circle(1,4,0.2),circle(2,9,0.2),graph(600,600,-10,10,-10,10,2x^2-x+3))}}}