How do you write the quadratic function in standard form given points (-1,5) (0,3) (3,9)?

1 Answer
Jan 19, 2017

#y=x^2-x+3#

Explanation:

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

We are told that solutions to the required quadratic are
#color(white)("XXX")(x,y) in {(-1,5),(0,3),(3,9)}#

That is
#{(5=a * (-1)^2+b * (-1) +c), (3=a * (0)^2+b * (0)+c), (9=a * (3)^2+b * (3) +c):}#

simplifying, we can write:
#{ ([1]color(white)("XX")a-b+c=5), ([2]color(white)("XXXXXXX")c=3), ([3]color(white)("XX")9a+3b+c=9) :}#

Using [2] we can further simplify [1] and [3] to get
#{ ([1]rarr[4]color(white)("XX")a-b=2), ([2]rarr[5]color(white)("XX")9a+3b=6color(white)("XX")rarrcolor(white)("XX")3a+b=2) :}#

Adding [4] and [5]
#{ color(white)("XXXX")[6]color(white)("XX")4a=4color(white)("XX")rarrcolor(white)("XX")a=1 :}#

Using [6], we can substitute #1# for #a# back in [4] to get
#{color(white)("XXXX")[7]color(white)("XX")1-b=2color(white)("XX")rarrcolor(white)("XX")b=-1 :}#

Therefore
#color(white)("XXX")(a,b)=(1,-1,3)#
and the required quadratic equation is
#color(white)("XXX")y=1x^2-1x+3 (=x^2-x+3)#