Question 460108
Since we're forcing a parabola to go through these three points, this means that when we plug in each 'x' into the general formula {{{f(x)=ax^2+bx+c}}}, we'll get 'y'. So essentially, we're plugging in corresponding values of 'x' and 'y' and forming equations, which we'll use to solve for 'a', 'b', and 'c'.



So for instance, if we plug in x=-1, then we'll get the output of y=-5. This means that


{{{5=a(-1)^2+b(-1)+c}}} which simplifies to {{{a-b+c=5}}}



Do the same with the second point to get the second equation {{{16a-4b+c=59}}}



Finally, do the same with the third point to get the third equation {{{9a-3b+c=37}}}


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After doing all that, we have the system



{{{system(a-b+c=5,16a-4b+c=59,9a-3b+c=37)}}}



which translates into this augmented matrix


{{{(matrix(3,4,1,-1,1,5,16,-4,1,59,9,-3,1,37))}}}



Row reduce this matrix (see <a href="http://i150.photobucket.com/albums/s91/jim_thompson5910/Algebra%20dot%20com/rref6-11-20112.png">this image</a> for full steps on how to do that) to get


{{{(matrix(3,4,1,0,0,2,0,1,0,-8,0,0,1,-5))}}}



Note: alternatively you can use a calculator to do the row reduction.



The last column represents the values of 'a', 'b', and 'c' respectively.



So {{{a=2}}}, {{{b=-8}}} and {{{c=-5}}}



So the quadratic equation that goes through the three given points is {{{y=2x^2-8x-5}}}



As a check, if we plug in x = -1, then....



{{{y=2(-1)^2-8(-1)-5}}}



{{{y=2(1)-8(-1)-5}}}



{{{y=2+8-5}}}



{{{y=5}}}



So plugging in x = -1 gives us y = 5, meaning that the point (-1,5) lies on the parabola. This partially verifies our answer.



Do the same with the other two points to verify this equation actually works.



Let me know if this helps.