Given three points (–1, 2), (0, 1), (1, –4) how do you write a quadratic function in standard form with the points?

1 Answer
May 17, 2016

#color(green)(y=-2x^2-3x+1)#

Explanation:

The generalized standard quadratic form is
#color(white)("XXX")ax^2+bx+c=y#

Substituting each of the given points: #(-1,2), (0,1), and (1,-4)#
gives three equations:
[1]#color(white)("XXX")a(-1)^2+b(-1)+c=2#
[2]#color(white)("XXX")a(0)^2+b(0)+c=1#
[3]#color(white)("XXX")a(1)^2+b(1)+c=0#

We can see directly from [2] that #c=1#

So [1] can be simplified as
[4]#color(white)("XXX")a-b+1=2#
and [3] can be simplified as
[5]#color(white)("XXX")a+b+1=-4#

Adding [4] and [5] gives
[6]#color(white)("XXX")2a+2=-2color(white)("XX")rarrcolor(white)("XX")a=-2#

Subtracting [4] from [5] gives
[7]#color(white)("XXX")2b=-6color(white)("XXXXX")rarrcolor(white)("XX")b=-3#

Substituting the calculated values for #a, b, and C# into the generalized quadratic form gives:
#color(white)("XXX")-2x^2-3x+1=y#