Question 1114193: Find the quadratic equation that fits the following set of data points.
(−2,−8),(−4,−12),(1,13)
f(x)=______.
(Use integers or fractions for any numbers in the expression. Do not factor.)
Found 2 solutions by Theo, MathTherapy:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the standard form of your quadratic equation is:
y = ax^2 + bx + c
when x = -2 and y = -8, this equation becomes:
-8 = 4a -2b + c
when x = -4, this equation becomes:
-12 = 16a - 4b + c
when x = 1, this equation becomes:
13 = a + b + c
you have 3 equations thst need to be solved simultaneously.
they are:
-8 = 4a - 2b + c
-12 = 16a - 4b + c
13 = a + b + c
you can reorder the terms to get:
4a - 2b + c = -8
16a - 4b + c = -12
a + b + c = 13
they are now in standard linear equation form.
when you solve these equations simultaneously, you will find that:
a = 1
b = 8
c = 4
since the standard form of your equation is y = ax^2 + bx + c, you get:
y = x^2 + 8x + 4
that's your quadratic equation.
when x = -2, you get y = (-2)^2 - 8*2 + 4 = -8
when x = -4, you get y = (-4)^2 - 8*4 + 4 = -12
when x = 1, you get y = 1^2 - 8*1 + 4 = 13
this confirms your equation is correct.
i'm assuming you know how to solve 3 equations in 3 unknowns simultaneously.
i solved as follows.
reference the worksheet shown below.
i started with equations 1, 2, and 3.
those are the circled numbers on the worksheet.
i then took equation 2 as is and multiplied equation 1 by 4 to get equation 4.
i then subtracted equation 4 from equation 2 to get equation 5.
i then took equation 2 as is and multiplied equation 3 by 16 to get equation 6.
i then subtracted equation 6 from equation 2 to get equation 7.
i then copied down equation 5 and 7 and proceeded to reduced these further so that i could solve for a single variable.
i multiplied equation 5 by 5 to get equation 8 and kept equation 7 as is.
i then subtracted equation 7 from equation 8 to get equation 9.
i then solved equation 9 for c to get c = 4
i then took equation 5 and replaced c with 4 and solved for b to get b = 8.
i then took equation 1 and replaced c with 4 and b with 8 and solved for a to get a = 1.
the result was a = 1, b = 8, c = 4.
that led to y = x^2 + 8x + 4, which is the solution shown above.
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
Find the quadratic equation that fits the following set of data points.
(−2,−8),(−4,−12),(1,13)
f(x)=______.
(Use integers or fractions for any numbers in the expression. Do not factor.)
Standard equation of a quadratic function: 
(− 2,− 8)
 ------ Substituting (- 2, - 8) for (x, y)
4a - 2b + c = - 8 ------- eq (i)
(− 4, − 12)
 ------ Substituting (- 4, - 12) for (x, y)
16a - 4b + c = - 12 ------- eq (ii)
(1, 13)
 ------ Substituting (1, 13) for (x, y)
a + b + c = 13 ------- eq (iii)
4a - 2b + c = - 8 ------- eq (i)
16a - 4b + c = - 12 ------- eq (ii)
a + b + c = 13 ------- eq (iii)
12a - 2b = - 4 ------ Subtracting eq (i) from eq (ii)
2(6a - b) = 2(- 2)
6a - b = - 2 ------ eq (iv)
3a - 3b = - 21 ------ Subtracting eq (iii) from eq (i)
3(a - b) = 3(- 7)
a - b = - 7 ------ eq (v)
5a = 5 ------- Subtracting eq (v) from eq (iv)
1 - b = - 7 ------- Substituting 1 for a in eq (v)
- b = - 7 - 1
- b = - 8
1 + 8 + c = 13 ------- Substituting 1 for a, and 8 for b in eq (iii)
9 + c = 13
 ------- Substituting 1 for a, 8 for b, and 4 for c
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