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. If using a calculator, you MUST upload picture of the calculator, or send it to your teacher in a webmail.
Note: Enter your answer and SHOW ALL THE STEPS NECESSARY to solve this problem in the space provided. You will not receive credit without all work shown.
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. If using a calculator, you MUST upload picture of the calculator, or send it to your teacher in a webmail.
Note: Enter your answer and SHOW ALL THE STEPS NECESSARY to solve this problem in the space provided. You will not receive credit without all work shown.
Answers
GPT-4o mini
To find a quadratic function \( P(d) = ad^2 + bd + c \) that represents the price of pizza \( P \) as a function of the diameter \( d \), we can use the three given points from the table.
The points we need are:
- \( (12, 15.80) \)
- \( (16, 22.20) \)
- \( (20, 35.00) \)
We can represent these points as equations by substituting \( d \) into the quadratic function:
1. For the first point \( (12, 15.80) \):
\[
15.80 = a(12)^2 + b(12) + c \implies 15.80 = 144a + 12b + c \quad \text{(Equation 1)}
\]
2. For the second point \( (16, 22.20) \):
\[
22.20 = a(16)^2 + b(16) + c \implies 22.20 = 256a + 16b + c \quad \text{(Equation 2)}
\]
3. For the third point \( (20, 35.00) \):
\[
35.00 = a(20)^2 + b(20) + c \implies 35.00 = 400a + 20b + c \quad \text{(Equation 3)}
\]
This gives us the following system of equations:
\[
\begin{align*}
144a + 12b + c &= 15.80 \quad \text{(1)} \\
256a + 16b + c &= 22.20 \quad \text{(2)} \\
400a + 20b + c &= 35.00 \quad \text{(3)}
\end{align*}
\]
Next, we will eliminate \( c \) by subtracting the equations in pairs:
### Step 1: Eliminate \( c \)
Subtract Equation 1 from Equation 2:
\[
(256a + 16b + c) - (144a + 12b + c) = 22.20 - 15.80
\]
This simplifies to:
\[
112a + 4b = 6.40 \quad \text{(Equation 4)}
\]
Subtract Equation 2 from Equation 3:
\[
(400a + 20b + c) - (256a + 16b + c) = 35.00 - 22.20
\]
This simplifies to:
\[
144a + 4b = 12.80 \quad \text{(Equation 5)}
\]
### Step 2: Solve for \( a \) and \( b \)
Now, we have a system of two equations in two variables:
\[
\begin{align*}
112a + 4b &= 6.40 \quad \text{(4)} \\
144a + 4b &= 12.80 \quad \text{(5)}
\end{align*}
\]
Next, we can eliminate \( b \) by subtracting Equation 4 from Equation 5:
\[
(144a + 4b) - (112a + 4b) = 12.80 - 6.40
\]
This gives:
\[
32a = 6.40 \implies a = \frac{6.40}{32} = 0.20
\]
### Step 3: Substitute \( a \) back to find \( b \)
Now substitute \( a = 0.20 \) into Equation 4:
\[
112(0.20) + 4b = 6.40
\]
\[
22.4 + 4b = 6.40
\]
\[
4b = 6.40 - 22.4
\]
\[
4b = -16.00 \implies b = -4.00
\]
### Step 4: Substitute \( a \) and \( b \) to find \( c \)
Now substitute \( a \) and \( b \) back into Equation 1 to find \( c \):
\[
144(0.20) + 12(-4.00) + c = 15.80
\]
\[
28.8 - 48 + c = 15.80
\]
\[
c = 15.80 + 48 - 28.8
\]
\[
c = 35.00 - 28.8 = 6.20
\]
### Conclusion
Now we have \( a \), \( b \), and \( c \):
- \( a = 0.20 \)
- \( b = -4.00 \)
- \( c = 6.20 \)
Thus, the equation for the price of the pizza as a function of its diameter is:
\[
P(d) = 0.20d^2 - 4.00d + 6.20
\]
The points we need are:
- \( (12, 15.80) \)
- \( (16, 22.20) \)
- \( (20, 35.00) \)
We can represent these points as equations by substituting \( d \) into the quadratic function:
1. For the first point \( (12, 15.80) \):
\[
15.80 = a(12)^2 + b(12) + c \implies 15.80 = 144a + 12b + c \quad \text{(Equation 1)}
\]
2. For the second point \( (16, 22.20) \):
\[
22.20 = a(16)^2 + b(16) + c \implies 22.20 = 256a + 16b + c \quad \text{(Equation 2)}
\]
3. For the third point \( (20, 35.00) \):
\[
35.00 = a(20)^2 + b(20) + c \implies 35.00 = 400a + 20b + c \quad \text{(Equation 3)}
\]
This gives us the following system of equations:
\[
\begin{align*}
144a + 12b + c &= 15.80 \quad \text{(1)} \\
256a + 16b + c &= 22.20 \quad \text{(2)} \\
400a + 20b + c &= 35.00 \quad \text{(3)}
\end{align*}
\]
Next, we will eliminate \( c \) by subtracting the equations in pairs:
### Step 1: Eliminate \( c \)
Subtract Equation 1 from Equation 2:
\[
(256a + 16b + c) - (144a + 12b + c) = 22.20 - 15.80
\]
This simplifies to:
\[
112a + 4b = 6.40 \quad \text{(Equation 4)}
\]
Subtract Equation 2 from Equation 3:
\[
(400a + 20b + c) - (256a + 16b + c) = 35.00 - 22.20
\]
This simplifies to:
\[
144a + 4b = 12.80 \quad \text{(Equation 5)}
\]
### Step 2: Solve for \( a \) and \( b \)
Now, we have a system of two equations in two variables:
\[
\begin{align*}
112a + 4b &= 6.40 \quad \text{(4)} \\
144a + 4b &= 12.80 \quad \text{(5)}
\end{align*}
\]
Next, we can eliminate \( b \) by subtracting Equation 4 from Equation 5:
\[
(144a + 4b) - (112a + 4b) = 12.80 - 6.40
\]
This gives:
\[
32a = 6.40 \implies a = \frac{6.40}{32} = 0.20
\]
### Step 3: Substitute \( a \) back to find \( b \)
Now substitute \( a = 0.20 \) into Equation 4:
\[
112(0.20) + 4b = 6.40
\]
\[
22.4 + 4b = 6.40
\]
\[
4b = 6.40 - 22.4
\]
\[
4b = -16.00 \implies b = -4.00
\]
### Step 4: Substitute \( a \) and \( b \) to find \( c \)
Now substitute \( a \) and \( b \) back into Equation 1 to find \( c \):
\[
144(0.20) + 12(-4.00) + c = 15.80
\]
\[
28.8 - 48 + c = 15.80
\]
\[
c = 15.80 + 48 - 28.8
\]
\[
c = 35.00 - 28.8 = 6.20
\]
### Conclusion
Now we have \( a \), \( b \), and \( c \):
- \( a = 0.20 \)
- \( b = -4.00 \)
- \( c = 6.20 \)
Thus, the equation for the price of the pizza as a function of its diameter is:
\[
P(d) = 0.20d^2 - 4.00d + 6.20
\]