Question 167529
The graph of {{{f(x) = ax^2 + bx + c}}} passes
through the points (1, k1), (2, k2), and (3, k3). Determine a,
b, and c for
<pre><font size = 4 color = "indigo"><b>
Plug {{{x=1}}} in:
{{{f(x) = ax^2 + bx + c}}}
{{{f(1) = a(1)^2 + b(1) + c}}}
{{{f(1) = a(1) + b(1) + c}}}
{{{f(1) = a+b+c}}}

Now since the graph passes through {{{matrix(1,5,"(",1,",",k[1],")")}}}
then {{{f(1)=k[1]}}}, so substitute {{{k[1]}}} for {{{f(1)}}} in

{{{f(1)=a+b+c}}}

or

{{{a+b+c=k[1]}}}

Plug {{{x=2}}} in:
{{{f(x) = ax^2 + bx + c}}}
{{{f(2) = a(2)^2 + b(2) + c}}}
{{{f(2) = a(4) + b(2) + c}}}
{{{f(2) = 4a+2b+c}}}

Now since the graph passes through {{{matrix(1,5,"(",2,",",k[2],")")}}}
then {{{f(2)=k[2]}}}, so substitute {{{k[2]}}} for {{{f(2)}}} in

{{{f(2)=4a+2b+c}}}

Plug {{{x=3}}} in:
{{{f(x) = ax^2 + bx + c}}}
{{{f(3) = a(3)^2 + b(3) + c}}}
{{{f(3) = a(9) + b(3) + c}}}
{{{f(3) = 9a+3b+c}}}

Now since the graph passes through {{{matrix(1,5,"(",3,",",k[3],")")}}}
then {{{f(3)=k[3]}}}, so substitute {{{k[3]}}} for {{{f(3)}}} in

{{{f(3) = 9a+3b+c}}}

{{{k[3] = 9a+3b+c}}} or

{{{9a+3b+c=k[3]}}}

So we have this system:


{{{system(a+b+c=k[1], 4a+2b+c=k[2], 9a+3b+c=k[3])}}}

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(A) {{{matrix(1,5,k1=-2,",",k2 = 1,",",k3 = 6)}}}

the system

{{{system(a+b+c=k[1], 4a+2b+c=k[2], 9a+3b+c=k[3])}}}

becomes

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

Solve this system and get {{{matrix(1,5, a=1,",", b=0, ",", c=-3)}}}

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(B) {{{matrix(1,5,k1=4,",",k2 = 3,",",k3 = -2)}}}

the system

{{{system(a+b+c=k[1], 4a+2b+c=k[2], 9a+3b+c=k[3])}}}

becomes

{{{system(a+b+c=4, 4a+2b+c=3, 9a+3b+c=-2)}}}

Solve this system and get {{{matrix(1,5, a=-2,",", b=5, ",", c=1)}}}

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(C) {{{matrix(1,5,k1=8,",",k2 = -5,",",k3 = 4)}}}

the system

{{{system(a+b+c=k[1], 4a+2b+c=k[2], 9a+3b+c=k[3])}}}

becomes

{{{system(a+b+c=8, 4a+2b+c=-5, 9a+3b+c=4)}}}

Solve this system and get {{{matrix(1,5, a=11,",", b=-46, ",", c=43)}}}

Edwin</pre>