Find a quadratic model for the set of values: (-2,-20), (0,-4), (4,-20). Show your work.
y = -3x^2 - 4x + 3
Is y = -3x^2 - 4x + 3 supposed to be your answer ?
Can't be right, since (0,-4) clearly does not satisfy your equation.
Method1:
let the equation be ax^2 + bx+c = y
for (0,-4), 0+0+c=-4 , c = -4
for (-2,-20), 4a-2b - 4 = -20
4a - 2b = -16
2a - b = -8
for (4,-20), 16a + 4b - 4 = -20
4a + b = -4
add the last two equations:
6a = -12
a = -2
back in 2a-b=-8
-4-b=-8
b = 4
so y = -2x^2 + 4x - 4
Method2:
the x of the vertex must lie half-way between -2 and 4 which would be 1
so we must have y = a(x-1)^2 + q
for (0,-4) ---> -4 = a+q
for (4,-20) --> -20 = 9a + q
subtract them:
-16 = 8a = -2
sub into a+q=-4
-2+q = -4
q = -2
equation: y = -2(x-1)^2 - 2
expand, if you must, to get the same equation as in Method 1
To find a quadratic model for the given set of values, we can assume the quadratic equation in the form of y = ax^2 + bx + c, where a, b, and c are constants.
Step 1: Substituting the first set of values (-2, -20) into the equation, we get:
-20 = a(-2)^2 + b(-2) + c
-20 = 4a - 2b + c
Step 2: Substituting the second set of values (0, -4) into the equation, we get:
-4 = a(0)^2 + b(0) + c
-4 = c
Therefore, we now have two equations:
Equation 1: -20 = 4a - 2b + c
Equation 2: c = -4
Step 3: Substituting the third set of values (4, -20) into Equation 1, we get:
-20 = 16a + 4b + c
Since c = -4 (from Equation 2), we can substitute it into Equation 1 and the above equation:
-20 = 4a - 2b - 4
-20 = 16a + 4b - 4
Simplifying both equations:
-4a + 2b = 16 <-- Equation 1
-16a - 4b = -16 <-- Equation 2
Step 4: Multiply Equation 1 by 2, then add it to Equation 2:
-8a + 4b = 32
-16a - 4b = -16
-------------
-24a = 16
Dividing both sides by -24 yields:
a = -2/3
Now that we know the value of a, we can substitute it back into Equation 1:
-4(-2/3) + 2b = 16
8/3 + 2b = 16
2b = 48/3 - 8/3
2b = 40/3
b = 20/3
Finally, we know the values of a and b, and c from Equation 2:
a = -2/3
b = 20/3
c = -4
Therefore, the quadratic model is:
y = (-2/3)x^2 + (20/3)x - 4