Question 1196721
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The response from the other tutor shows a typical formal algebraic solution.<br>
Since your post gives no indication of your level of mathematical studies, I will demonstrate a very different method for solving the problem.<br>
Since a quadratic model is specified, consider a sequence of consecutive terms of a quadratic sequence.  We are given the values for x=-1, x=2, and x=3; use variables to represent the terms of the sequence for x=0 and x=1:<br>
10, a, b, 4, -6<br>
The method of finite differences (do an internet search if you are interested in learning more about it) tells us that, because the model is quadratic (a second degree polynomial), the second differences of the sequence are constant.  Find the second differences of the sequence and use the fact that they are equal to find a and b.<br><pre>

    10      a         b        4     -6  <-- the consecutive terms of the sequence
      a-10       b-a      4-b    -10     <-- the first differences
          b-2a+10   a-2b+4   b-14        <-- the second differences</pre>
{{{b-2a+10=b-14}}}
{{{-2a+10=-14}}}
{{{24=2a}}}
{{{a=12}}}<br>
{{{b-14=a-2b+4=12-2b+4}}}
{{{b-14=16-2b}}}
{{{3b=30}}}
{{{b=10}}}<br>
The quadratic sequence is<br>
10, 12, 10, 4, -6<br>
The set of coordinates corresponding to the sequence are<br>
(-1,10), (0,12), (1,10), (2,4), and (3,-6)<br>
The identical y values for x=-1 and x=1 tell us the vertex of the graph of the function is at (0,12), so the function is of the form<br>
{{{y=ax^2+12}}}<br>
Use any of the other known points to determine (by whatever method you want) that a = -2.<br>
Then the quadratic model for the given set of 3 points is<br>
ANSWER: {{{y=-2x^2+12}}}<br>