SOLUTION: Find the quadratic in standard form that goes through (-4,27), (1,2) (3,34)
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Question 328470: Find the quadratic in standard form that goes through (-4,27), (1,2) (3,34)
Found 2 solutions by solver91311, ankor@dixie-net.com:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
\ =\ ax^2\ +\ bx +\ c)
\ =\ a(-4)^2\ +\ b(-4) +\ c\ =\ 27)
\ =\ a(1)^2\ +\ b(1) +\ c\ =\ 2)
\ =\ a(3)^2\ +\ b(3) +\ c\ =\ 34)
Solve the system for your coefficients and plug them into:
John
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Find the quadratic in standard form that goes through (-4,27), (1,2) (3,34)
:
Using the form, ax^2 + bx + c = y, write an equation for each pair
:
x=-4, y=27
a(-4^2) + b(-4) + c = 27
16a - 4b + c = 27
Do the same with the other two pairs
x=1, y=2
a + b + c = 1
:
x=3, y=34
9a + 3b + c = 34
:
Subtract the 2nd equation from the 1st equation
16a - 4b + c = 27
a + b + c = 2
--------------------eliminates c
15a - 5b = 25
simplify divide by 5
3a - b = 5
:
Subtract the 2nd equation from the 3rd equation
9a + 3b + c = 34
a + b + c = 2
--------------------Eliminates c
8a + 2b = 32
simplify, divide by 2
4a + b = 16
:
Two equation which we can handle easily with elimination again
3a - b = 5
4a + b = 16
---------------adding eliminates b, find a
7a = 21
a = 21/7
a = 3
:
Find b using the equation 4a + b = 16,
4(3) + b = 16
12 + b = 16
b = 16 - 12
b = 4
:
Find c using the equation a + b + c = 2
3 + 4 + c = 2
c = 2 - 7
c = -5
:
The equation: y = 3x^2 + 4x - 5
:
You can prove this to yourself by substituting for x and finding y for each pair
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