Question 746893
{{{Y=Ax^2 - Bx + C}}}

if the quadratic function which takes the value {{{41}}} at {{{x=-2}}}, then


{{{41=A(-2)^2 - B(-2) + C}}}

{{{41=4A + 2B + C}}}.......eq.1

and if it  takes the value {{{20}}} at {{{x=5}}} 

{{{20=A(5)^2 - B(5) + C}}}

{{{20=25A -5B + C}}}.......eq.2

and is minimized at {{{x=21}}}

{{{0=A*21^2 - B*21 + C}}}

{{{0=441A - 21B + C}}}......eq.3


solve the system:


{{{41=4A + 2B + C}}}.......eq.1

{{{20=25A -5B + C}}}.......eq.2

{{{0=441A - 21B + C}}}......eq.3
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{{{41=4A + 2B + C}}}.......eq.1...solve for {{{C}}}

{{{41-4A -2B=C}}}......substitute in eq.2a


{{{20=25A -5B + (41-4A -2B)}}}.......eq.2

{{{20=25A -5B + 41-4A -2B}}}

{{{20-41=21A -7B }}}.......solve for {{{B}}}

{{{-21=21A -7B }}}

{{{7B=21A +21 }}}

{{{7B/7=21A/7 +21/7 }}}

{{{B=3A +3 }}}.............substitute in eq.2a



{{{41-4A -2B=C}}}........eq.2a

{{{41-4A -2(3A +3)=C}}}

{{{41-4A -6A -6=C}}}

{{{35-10A=C}}}........eq.2b

now use {{{0=441A - 21B + C}}}......eq.3 and substitute {{{B}}} and {{{C}}}

{{{0=441A - 21(3A +3) + (35-10A)}}}......eq.3....solve for {{{A}}}

{{{0=441A - 63A -63 + 35-10A}}}

{{{0=368A  -63 + 35}}}

{{{0=368A  -28}}}

{{{28=368A }}}

{{{A=28/368 }}}

{{{highlight(A=0.076) }}}

now find {{{B}}} and {{{C}}}

{{{B=3A +3 }}}

{{{B=3*0.076 +3 }}}

{{{highlight(B=3.228) }}}

{{{35-10A=C}}}

{{{35-10*0.076=C}}}

{{{35-0.76=C}}}

{{{highlight(34.24=C)}}}