Question 1102896
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n points can be fitted with a polynomial of degree (n-1).  With 4 given points, you need a polynomial of degree 3:
{{{P(x) = ax^3+bx^2+cx+d}}}<br>
Simply make 4 equations in the unknown coefficients a, b, c, and d using the coordinates of the given points.
(2,10) --> {{{8a+4b+2c+d = 10}}}  [1]
(-1,-4) --> {{{-a+b-c+d = -4}}}  [2]
(0,2) --> {{{d = 2}}}  [3]
(1,-2) --> {{{a+b+c+d = -2}}}  [4]<br>
Use [3] to simplify the others:
{{{8a+4b+2c = 8}}}  [5]
{{{-a+b-c = -6}}}  [6]
{{{a+b+c = -4}}}  [7]<br>
Add [6] and [7] to solve for b:
{{{2b = -10}}}
{{{b = -5}}}  [8]<br>
Use [8] to simplify [5] and [7]:
{{{8a-20+2c = 8}}}
{{{8a+2c = 28}}}
{{{4a+c = 14}}}  [9]
{{{a-5+c = -4}}}
{{{a+c = 1}}}  [10]<br>
Solve [9] and [10] by elimination to solve for a and c:
{{{3a = 13}}}
{{{a = 13/3}}}
{{{c = -10/3}}}<br>
The polynomial is
{{{p(x) = (13/3)x^3-5x^2-(10/3)x+2}}}<br>
Here is the graph...<br<
{{{graph(200,400,-3,3,-5,11,(13/3)x^3-5x^2-(10/3)x+2)}}}