SOLUTION: write a quadratic function in standard form that goes through the given points (1,3) (2,1) (-2,-15)

Question 118256: write a quadratic function in standard form that goes through the given points (1,3) (2,1) (-2,-15)
Found 2 solutions by stanbon, ankor@dixie-net.com:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
write a quadratic function in standard form that goes through the given points (1,3) (2,1) (-2,-15)
---------------
Quadratic Form:
y = ax^2 + bx + c
Substitute each pair of x/y values then solve for a,b,c
a + b +c = 3
4a+2b +c = 1
4a-2b +c = -15
------------------
I used the Matrix function on a TI calculator to get:
a = -2
b = 4
c = 1
---------
EQUATION:
y = -2x^2+4x+1
===================
Cheers,
Stan H.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
If you can't don't want to use the matrix method, you also use the elimination method
:
write a quadratic function in standard form that goes through the given points (1,3) (2,1) (-2,-15)
:
In form ax^2 + bx + c = y; solve for a, b, c
:
1,3: eq1: a + b + c = 3
2,1: eq2: 4a + 2b + c = 1
-2,-15: eq3: 4a - 2b + c = -15
:
Using equation 2 & 3 we can find b:
4a + 2b + c = 1
4a - 2b + c = -15
------------------subtracting eliminates a and c
0a + 4b + 0x = 16
4b = 16
b = 16/4
b = 4
:
Substitute 4 for b in eq1 & 2
a + 4 + c = 3
a + c = 3 - 4
a + c = -1
and
4a + 2(4) + c = 1
4a + 8 + c = 1
4a + c = 1 - 8
4a + c = -7
:
Using these two equations:
4a + c = -7
a + c = -1
-------------subtracting eliminates c
3a + 0c = -6
3a = -6
a = -6/3
a = -2
:
Find c, by substituting for a & b in a + b + c = 3
-2 + 4 + c = 3
c + 2 = 3
c = 3 - 2
c = 1
:
Therefore we have y = -2x^2 + 4x + 1
;
Is this what you had in mind?
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