If the graph of #f(x) = ax^3+bx^2+cx+d# passes through the points #(0,10)#, #(1,7)#, #(3,-11)# and #(4,-24)# then what are the values of #a, b, c, d# ?

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Note that given points #(x_1, y_1)#, #(x_2, y_2)#, #(x_3, y_3)#, #(x_4, y_4)# (with #x_1, x_2, x_3, x_4# all distinct) then the formula of a cubic polynomial passing through them can be written:

#((x-x_2)(x-x_3)(x-x_4))/((x_1-x_2)(x_1-x_3)(x_1-x_4))y_1 + ((x-x_1)(x-x_3)(x-x_4))/((x_2-x_1)(x_2-x_3)(x_2-x_4))y_2+((x-x_1)(x-x_2)(x-x_4))/((x_3-x_1)(x_3-x_2)(x_3-x_4))y_3+((x-x_1)(x-x_2)(x-x_3))/((x_4-x_1)(x_4-x_2)(x_4-x_3))y_4#

That is a little tedious to calculate, so let's use a different approach for the given example.

Suppose the cubic also passes through #(2, p)#. Then the samples #(0, 10)#, #(1, 7)#, #(2, p)#, #(3, -11)#, #(4, -14)# are evenly spaced along the #x# axis, with each step being #1#.

Hence we can analyse the #y# values as a sequence using the method of differences.

Write out the initial sequence:

#color(blue)(10), 7, p, -11, -14#

Write out the sequence of differences between successive terms:

#color(blue)(-3), p-7, -p-11, -3#

Hmmm. Notice that the first and last differences are both #-3# and we can make the two centre differences the same by putting #p=-2#, so let's do that to get:

#color(blue)(-3), -9, -9, -3#

Then the sequence of differences of those differences is:

#color(blue)(-6), 0, 6#

The sequence of differences of those differences is:

#color(blue)(6), 6#

Having arrived at a constant sequence (and incidentally verified our choice of #p#), we can use the initial term of each of the sequences we found to make coefficients of the required cubic:

#f(x) = color(blue)(10)/(0!)+color(blue)(-3)/(1!)x+color(blue)(-6)/(2!)x(x-1)+color(blue)(6)/(3!)x(x-1)(x-2)#

#color(white)(f(x)) = 10-3x-3x^2+3x+x^3-3x^2+2x#

#color(white)(f(x)) = x^3-6x^2+2x+10#

So #a=1#, #b=-6#, #c=2# and #d=10#