How do you write an equation of the quadratic function whose graph contains the points (-4,-57), (2, 39), (5, 168)?

General form of a quadratic equation is
#color(white)("XX")y=ax^2+bx+c#

So for the given points, we have
[1]#color(white)("XX")-57=a(-4)^2+b(-4)+c#
[2]#color(white)("XX")39=a(2)^2+b(2)+c#
[3]#color(white)("XX")168=a(5)^2+b(5)+c#

Simplifying these:
[3]#color(white)("XX")16a-4b+c=-57#
[4]#color(white)("XX")4a+2b+c=39#
[5]#color(white)("XX")25a+5b+c=168#

Subtracting [4] from [3]
[6]#color(white)("XX")12a-6b= -96#
Subtracting [4] from [5]
[7]#color(white)("XX")21a+3b=129#

Multiplying [7] by 2
[8]#color(white)("XX")42a+6b=258#

Adding [6] and [8]
[9]#color(white)("XX")54a = 162#

Dividing [9] by 3
[10]#color(white)("XX")a=3#

Substituting #3# for #a# in [6]
[11]#color(white)("XX")12(3)-6b=-96#

Simplifying
[12]#color(white)("XX")-6b =-132#

Dividing [12] by #-6#
[13]#color(white)("XX")b=12#

Substituting #12# for #b# (from [13]) and #3# for #a# (from [10]) in [4]
[14]#color(white)("XX")4(3)+2(12)+c = 39#

Simplifying
[15]#color(white)("XX")12+24+c=39#

[16]#color(white)("XX")c=3#

First, we will create three equations, one for each point. It will be in the general form #y=ax^2+bx+c#.

First Equation from P(-4,-57)
#y=ax^2+bx+c#
#(-57)=a(-4)^2+b(-4)+c#
#-57=16a-4b+c#

Second Equation from P(2,39)
#y=ax^2+bx+c#
#(39)=a(2)^2+b(2)+c#
#39=4a+2b+c#

Third Equation from P(5,168)
#y=ax^2+bx+c#
#(168)=a(5)^2+b(5)+c#
#168=25a+5b+c#

System of Equations:
#-57=16a-4b+c#
#39=4a+2b+c#
#168=25a+5b+c#

Now, we can solve for the values of #a#, #b#, and #c#. Let's try getting two more equations to help us solve this system.

Fourth Equation taken from the first and second equations:
We will first get the value of #c# in the first equation then we will substitute it into #c# in the second equation.
#-57=16a-4b+c#
#-57-16a+4b=16a-4b+c-16a+4b#
#-57-16a+4b=c#

#39=4a+2b+c#
#39=4a+2b+(-57-16a+4b)#
#39=4a+2b-57-16a+4b#
#39+57=-12a+6b#
#96=-12a+6b#

Let's simplify this to:
#16=-2a+b#

Fifth Equation taken from the first and third equations:
We will substitute the value of #c# in the first equation into #c# in the third equation.
#168=25a+5b+c#
#168=25a+5b+(-57-16a+4b)#
#168=25a+5b-57-16a+4b#
#168+57=9a+9b#
#225=9a+9b#

Let's simplify this to:
#25=a+b#

Solution
Now, we will use the 4th and 5th equation to solve for #a# and #b#. I'll use the elimination method since it is easier:

#-2a+b=16#
#ul(-a-b=-25)#
#-3a=-9#
#a=3#

#25=a+b#
#25=3+b#
#b=22#

To solve for #c# just substitute the values of #a# and #b# into any of the first three equations.
#39=4a+2b+c#
#39=4(3)+2(22)+c#
#39=12+44+c#
#c=-17#

So the values of #a#, #b#, and #c# are:
#a=3#
#b=22#
#c=-17#

Finally, to get the quadratic equation, just substitute the values of #a#, #b#, and #c# into the general form #y=ax^2+bx+c#.
#y=ax^2+bx+c#
#color(red)(y=3x^2+22x-17)#