How do you write a quadratic function in standard form whose graph passes through points (2,4), (3,7), (1,3)?
2 Answers
Explanation:
Here's one method...
Putting the points in increasing order of
#(1, 3)# ,#(2, 4)# ,#(3, 7)#
Note that the
Hence we can analyse this as a sequence of
Write down the sequence of
#color(blue)(3), 4, 7#
Write down the sequence of differences between pairs of consecutive terms:
#color(blue)(1), 3#
Write down the sequence of differences of differences:
#color(blue)(2)#
Having reached a constant sequence (albeit of just one term), we can write a general formula for a term of the initial sequence of
#a_n = color(blue)(3)/(0!)+color(blue)(1)/(1!)(n1)+color(blue)(2)/(2!)(n1)(n2)#
#color(white)(a_n) = 3n+1n^2+3n2#
#color(white)(a_n) = n^2+2n4" "# for#n = 1,2,3#
Hence our function can be written:
#f(x) = x^2+2x4#
Explanation:
Here's a general purpose method for matching a polynomial to a set of points...
The formula
Given a set of points:
#{ (x_1, y_1),...,(x_n, y_n) }#
with distinct values
A polynomial that passes through them is:
#((xx_2)(xx_3)...(xx_n))/((x_1x_2)(x_1x_3)...(x_1x_n))y_1 +#
#((xx_1)(xx_3)...(xx_n))/((x_2x_1)(x_2x_3)...(x_2x_n))y_2 + ...#
#+ ((xx_1)(xx_2)...(xx_(n1)))/((x_nx_1)(x_nx_2)...(x_nx_(n1)))y_n#
#= sum_(k=1)^n (prod_(j = 1, j != k)^n (xx_j)/(x_kx_j)) y_k#
How does it work?
That may look rather complicated, but notice how it works:
Looking at the first expression:
#((xx_2)(xx_3)...(xx_n))/((x_1x_2)(x_1x_3)...(x_1x_n)) = prod_(j = 2)^n (xx_j)/(x_1x_j)#
Note that:

It takes the value
#1# when#x=x_1# since the numerator and denominator become identical. 
It takes the value
#0# when#x=x_k# for any#k# apart from#1# , since one of the factors in the numerator will be#0# . 
It is a polynomial of degree
#n1# .
So when we multiply this by
Summing this with the similar polynomials for
Application
In the given example, with points:
#{ ((x_1, y_1) = (2, 4)), ((x_2, y_2) = (3, 7)), ((x_3, y_3) = (1, 3)) :}#
We can write:
#f(x) = ((x3)(x1))/((23)(21))(4)+((x2)(x1))/((32)(31))(7)+((x2)(x3))/((12)(13))(3)#
#color(white)(f(x)) = 4(x^24x+3)7/2(x^23x+2)3/2(x^25x+6)#
#color(white)(f(x)) = x^2+2x4#