Question
At Pizza Palace, the price of a pizza is a quadratic function of the diameter. The prices of three pizzas at Pizza Palace are shown in the following table: Diameter of Pizza (inches) Price of Pizza (dollars) 12" $15.80 16" $22.20 20" $35.00 Question Use the three points from the table to write the equation of the parabola that passes through the points.
Answers
GPT-4o mini
We have three points corresponding to the given diameters and prices of pizzas at Pizza Palace:
1. \( (12, 15.80) \)
2. \( (16, 22.20) \)
3. \( (20, 35.00) \)
We can express the price \( P \) as a quadratic function of the diameter \( d \) in the form:
\[
P(d) = ad^2 + bd + c
\]
where \( a \), \( b \), and \( c \) are coefficients we need to determine.
Using the three points, we can set up a system of equations:
1. From \( (12, 15.80) \):
\[
a(12^2) + b(12) + c = 15.80
\]
\[
144a + 12b + c = 15.80 \quad \text{(Equation 1)}
\]
2. From \( (16, 22.20) \):
\[
a(16^2) + b(16) + c = 22.20
\]
\[
256a + 16b + c = 22.20 \quad \text{(Equation 2)}
\]
3. From \( (20, 35.00) \):
\[
a(20^2) + b(20) + c = 35.00
\]
\[
400a + 20b + c = 35.00 \quad \text{(Equation 3)}
\]
Now we will organize our equations:
1. \( 144a + 12b + c = 15.80 \)
2. \( 256a + 16b + c = 22.20 \)
3. \( 400a + 20b + c = 35.00 \)
Next, we will eliminate \( c \) from these equations. We can do this by subtracting Equation 1 from Equation 2, and Equation 2 from Equation 3.
Subtract Equation 1 from Equation 2:
\[
(256a + 16b + c) - (144a + 12b + c) = 22.20 - 15.80
\]
\[
(256 - 144)a + (16 - 12)b = 6.40
\]
\[
112a + 4b = 6.40 \quad \text{(Equation 4)}
\]
Subtract Equation 2 from Equation 3:
\[
(400a + 20b + c) - (256a + 16b + c) = 35.00 - 22.20
\]
\[
(400 - 256)a + (20 - 16)b = 12.80
\]
\[
144a + 4b = 12.80 \quad \text{(Equation 5)}
\]
Now we have a new system of two equations with two unknowns:
4. \( 112a + 4b = 6.40 \)
5. \( 144a + 4b = 12.80 \)
We can now eliminate \( b \) by subtracting Equation 4 from Equation 5:
\[
(144a + 4b) - (112a + 4b) = 12.80 - 6.40
\]
\[
32a = 6.40
\]
\[
a = \frac{6.40}{32} = 0.20
\]
Now we can substitute \( a = 0.20 \) back into Equation 4 to find \( b \):
\[
112(0.20) + 4b = 6.40
\]
\[
22.40 + 4b = 6.40
\]
\[
4b = 6.40 - 22.40
\]
\[
4b = -16
\]
\[
b = -4
\]
Now that we have \( a \) and \( b \), we can find \( c \) by substituting \( a \) and \( b \) back into one of the original equations; we can use Equation 1:
\[
144(0.20) + 12(-4) + c = 15.80
\]
\[
28.8 - 48 + c = 15.80
\]
\[
c = 15.80 - 28.8 + 48
\]
\[
c = 15.80 - 28.8 + 48 = 35.00
\]
Thus, we have found:
- \( a = 0.20 \)
- \( b = -4 \)
- \( c = -35.00 \)
Putting it all together, the equation of the parabola that models the price of pizza as a function of its diameter is:
\[
P(d) = 0.20d^2 - 4d + 35.00
\]
1. \( (12, 15.80) \)
2. \( (16, 22.20) \)
3. \( (20, 35.00) \)
We can express the price \( P \) as a quadratic function of the diameter \( d \) in the form:
\[
P(d) = ad^2 + bd + c
\]
where \( a \), \( b \), and \( c \) are coefficients we need to determine.
Using the three points, we can set up a system of equations:
1. From \( (12, 15.80) \):
\[
a(12^2) + b(12) + c = 15.80
\]
\[
144a + 12b + c = 15.80 \quad \text{(Equation 1)}
\]
2. From \( (16, 22.20) \):
\[
a(16^2) + b(16) + c = 22.20
\]
\[
256a + 16b + c = 22.20 \quad \text{(Equation 2)}
\]
3. From \( (20, 35.00) \):
\[
a(20^2) + b(20) + c = 35.00
\]
\[
400a + 20b + c = 35.00 \quad \text{(Equation 3)}
\]
Now we will organize our equations:
1. \( 144a + 12b + c = 15.80 \)
2. \( 256a + 16b + c = 22.20 \)
3. \( 400a + 20b + c = 35.00 \)
Next, we will eliminate \( c \) from these equations. We can do this by subtracting Equation 1 from Equation 2, and Equation 2 from Equation 3.
Subtract Equation 1 from Equation 2:
\[
(256a + 16b + c) - (144a + 12b + c) = 22.20 - 15.80
\]
\[
(256 - 144)a + (16 - 12)b = 6.40
\]
\[
112a + 4b = 6.40 \quad \text{(Equation 4)}
\]
Subtract Equation 2 from Equation 3:
\[
(400a + 20b + c) - (256a + 16b + c) = 35.00 - 22.20
\]
\[
(400 - 256)a + (20 - 16)b = 12.80
\]
\[
144a + 4b = 12.80 \quad \text{(Equation 5)}
\]
Now we have a new system of two equations with two unknowns:
4. \( 112a + 4b = 6.40 \)
5. \( 144a + 4b = 12.80 \)
We can now eliminate \( b \) by subtracting Equation 4 from Equation 5:
\[
(144a + 4b) - (112a + 4b) = 12.80 - 6.40
\]
\[
32a = 6.40
\]
\[
a = \frac{6.40}{32} = 0.20
\]
Now we can substitute \( a = 0.20 \) back into Equation 4 to find \( b \):
\[
112(0.20) + 4b = 6.40
\]
\[
22.40 + 4b = 6.40
\]
\[
4b = 6.40 - 22.40
\]
\[
4b = -16
\]
\[
b = -4
\]
Now that we have \( a \) and \( b \), we can find \( c \) by substituting \( a \) and \( b \) back into one of the original equations; we can use Equation 1:
\[
144(0.20) + 12(-4) + c = 15.80
\]
\[
28.8 - 48 + c = 15.80
\]
\[
c = 15.80 - 28.8 + 48
\]
\[
c = 15.80 - 28.8 + 48 = 35.00
\]
Thus, we have found:
- \( a = 0.20 \)
- \( b = -4 \)
- \( c = -35.00 \)
Putting it all together, the equation of the parabola that models the price of pizza as a function of its diameter is:
\[
P(d) = 0.20d^2 - 4d + 35.00
\]