Find a quadratic model for the set of values: (-2, -20), (0, -4), (4, -20). Show your work

Please help!

Did you notice that two of the points have the same y value of -20

so the vertex must lie on x = 1 midway between the two x's , (because (-2 + 4)/2) = 1 )
So, let the quadratic be
y = a(x-1)^2 + c
for the point (0,-4)
-4 = a + c
for the point (4,-20)
-20 = 9a + c
subtract them:
-16 = 8a
a = -2
plug into -4=a+c
-4 = -2+c ----> c = -2

the quadratic is : y = -2(x-1)^2 - 2

To find a quadratic model for the given set of values, we need to follow a few steps:

Step 1: Formulate the quadratic equation.
The quadratic equation in the general form is given by:
𝑦 = 𝑎𝑥^2 + 𝑏𝑥 + 𝑐

Step 2: Substitute the given points to create a system of equations.
Using the given points (-2, -20), (0, -4), and (4, -20), we can substitute the x and y values into the quadratic equation to form three equations.

For (-2, -20):
(-20) = 𝑎(-2)^2 + 𝑏(-2) + 𝑐

For (0, -4):
(-4) = 𝑎(0)^2 + 𝑏(0) + 𝑐

For (4, -20):
(-20) = 𝑎(4)^2 + 𝑏(4) + 𝑐

Step 3: Solve the system of equations.
Now, we can solve this system of equations to find the values of 𝑎, 𝑏, and 𝑐.

Using the first equation, we get:
(-20) = 4𝑎 - 2𝑏 + 𝑐

Using the second equation, we get:
(-4) = 𝑐

Using the third equation, we get:
(-20) = 16𝑎 + 4𝑏 + 𝑐

Substituting the value of 𝑐 from the second equation into the first and third equations, we have:
(-20) = 4𝑎 - 2𝑏 - 4 ... (Equation 1)
(-20) = 16𝑎 + 4𝑏 - 4 ... (Equation 2)

Now, let's solve this system of equations using any method you prefer, such as substitution or elimination.

Using the substitution method, we can solve Equation 1 for 𝑐:
𝑐 = -4

Substituting this value of 𝑐 into Equation 2, we get:
(-20) = 16𝑎 + 4𝑏 - 4
(-16) = 16𝑎 + 4𝑏

Dividing Equation 2 by 4, we have:
(-4) = 4𝑎 + 𝑏

Now, we have the following system of equations:
4𝑎 - 2𝑏 = -16
4𝑎 + 𝑏 = -4

Using the elimination method, we can add Equation 2 to Equation 1:
(4𝑎 - 2𝑏) + (4𝑎 + 𝑏) = (-16) + (-4)
8𝑎 - 𝑏 = -20 ... (Equation 3)

Now, we have the following system of equations:
4𝑎 - 2𝑏 = -16
8𝑎 - 𝑏 = -20

Multiplying Equation 1 by 2, we have:
8𝑎 - 4𝑏 = -32

Adding Equation 3 to Equation 1, we get:
8𝑎 - 𝑏 + 8𝑎 - 4𝑏 = -20 + (-32)
16𝑎 - 5𝑏 = -52 ... (Equation 4)

Now, we have the following system of equations:
8𝑎 - 4𝑏 = -32
16𝑎 - 5𝑏 = -52

Using the elimination method again, we can multiply Equation 1 by 2 and subtract it from Equation 4:

(16𝑎 - 5𝑏) - (16𝑎 - 4𝑏) = (-52) - (-32)
16𝑎 - 5𝑏 -16𝑎 + 4𝑏 = -52 + 32
-𝑏 = -20

Dividing both sides by -1, or multiplying both sides by -1, we get:
𝑏 = 20

Substituting this value of 𝑏 into Equation 2, we have:
4𝑎 + 20 = -4
4𝑎 = -24
𝑎 = -6

Now, we have found the values of 𝑎 = -6, 𝑏 = 20, and 𝑐 = -4.

Step 4: Write the quadratic model.
Substituting these values into the general quadratic equation:
𝑦 = (-6)𝑥^2 + (20)𝑥 + (-4)

Thus, the quadratic model for the given set of values is:
𝑦 = -6𝑥^2 + 20𝑥 - 4