THIS IS MY FINAL EXAM QUESTION PLEASE HELP ME SO I CAN PASS IT (((
1 answer:
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Answers:
- -1 in the first box
- 19/2 or 9.5 in the second box
- -12 in the third box
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Work Shown:
We'll use the template y = ax^2 + bx + c
- If (x,y) = (2,3) is on the parabola, then y = ax^2+bx+c turns into 3 = a(2)^2 + b(2) + c. That simplifies to 4a+2b+c = 3
- If (x,y) = (6,9) is on the parabola, then y = ax^2+bx+c turns into 9 = a(6)^2 + b(6) + c. That simplifies to 36a+6b+c = 9
- If (x,y) = (8,0) is on the parabola, then y = ax^2+bx+c turns into 0 = a(8)^2 + b(8) + c. That simplifies to 64a+8b+c = 0
The system of equations is
We have 3 equations and 3 unknowns.
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Let's solve the first equation for c
4a+2b+c = 3
2b+c = 3-4a
c = 3-4a-2b
c = -4a-2b+3
Plug that into the second equation and simplify
36a+6b+c = 9
36a+6b+(-4a-2b+3) = 9
32a+4b+3 = 9
32a+4b = 9-3
32a+4b = 6
2(16a+2b) = 6
16a+2b = 6/2
16a+2b = 3 .... we'll use this later
Plug that value of c into the third equation as well
64a+8b+c = 0
64a+8b+(-4a-2b+3) = 0
60a+6b+3 = 0
3(20a+2b+1) = 0
20a+2b+1 = 0
20a+2b = -1
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We have a new system of equations. This time it deals with 2 variables instead of 3. We can think of this as a reduced equivalent system.
Note how subtracting the terms straight down has the b terms canceling (since 2b-2b = 0b = 0)
The 'a' terms subtract to 16a-20a = -4a
The terms on the right hand side subtract to 3-(-1) = 3+1 = 4
We end up with the equation -4a = 4 which solves to a = -1
Use this value of 'a' to find b
16a+2b = 3
16(-1)+2b = 3
-16+2b = 3
2b = 3+16
2b = 19
b = 19/2
b = 9.5
Now use the values of a and b to find c
4a+2b+c = 3
4(-1)+2(9.5)+c = 3
-4+19+c = 3
15+c = 3
c = 3-15
c = -12
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Summary:
The values of a,b,c we found were
- a = -1
- b = 19/2 = 9.5
- c = -12
So the function is
f(x) = -x^2 + (19/2)x - 12
which is equivalent to
f(x) = -x^2 + 9.5x - 12
To check this answer, plug in x = 2, x = 6, x = 8 one at a time. You should get y = 3, y = 9 and y = 0 in that order.