y=ax^{2}+bx+c using (0,325) (50,300) ( 75,250)

To find the quadratic equation \( y = ax^2 + bx + c \) that passes through the points \((0, 325)\), \((50, 300)\), and \((75, 250)\), we need to solve for the coefficients \(a\), \(b\), and \(c\). 

### Step 1: Substitute the points into the equation
For each point \((x, y)\), substitute the values into the equation:

1. For \((0, 325)\):
   \[
   325 = a(0)^2 + b(0) + c \implies c = 325
   \]
   
2. For \((50, 300)\):
   \[
   300 = a(50)^2 + b(50) + 325
   \]
   Simplifying:
   \[
   300 = 2500a + 50b + 325 \implies 2500a + 50b = -25
   \]
   Dividing by 25:
   \[
   100a + 2b = -1 \quad \text{(Equation 1)}
   \]

3. For \((75, 250)\):
   \[
   250 = a(75)^2 + b(75) + 325
   \]
   Simplifying:
   \[
   250 = 5625a + 75b + 325 \implies 5625a + 75b = -75
   \]
   Dividing by 25:
   \[
   225a + 3b = -3 \quad \text{(Equation 2)}
   \]

### Step 2: Solve the system of linear equations
We now have two equations with two variables:

\[
100a + 2b = -1 \quad \text{(Equation 1)}
\]
\[
225a + 3b = -3 \quad \text{(Equation 2)}
\]

Let's solve this system using substitution or elimination.

1. **Multiply Equation 1 by 3**:
   \[
   300a + 6b = -3
   \]
   
2. **Multiply Equation 2 by 2**:
   \[
   450a + 6b = -6
   \]

Now, subtract the modified Equation 1 from the modified Equation 2:

\[
(450a + 6b) - (300a + 6b) = -6 - (-3)
\]
Simplifying:
\[
150a = -3 \implies a = -\frac{1}{50}
\]

### Step 3: Find \(b\)
Substitute \( a = -\frac{1}{50} \) into Equation 1:

\[
100\left(-\frac{1}{50}\right) + 2b = -1
\]
\[
-2 + 2b = -1 \implies 2b = 1 \implies b = \frac{1}{2}
\]

### Step 4: Write the final equation
Now that we have \(a = -\frac{1}{50}\), \(b = \frac{1}{2}\), and \(c = 325\), the quadratic equation is:

\[
y = -\frac{1}{50}x^2 + \frac{1}{2}x + 325
\]

### Summary
The quadratic equation that passes through the points \((0, 325)\), \((50, 300)\), and \((75, 250)\) is:

\[
y = -\frac{1}{50}x^2 + \frac{1}{2}x + 325
\]

Would you like further details or have any questions?

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**Related Questions:**

1. How do you derive the general form of a quadratic equation from three points?
2. What is the vertex form of the quadratic equation obtained?
3. How can you determine the axis of symmetry for the given quadratic equation?
4. What is the significance of the coefficients \(a\), \(b\), and \(c\) in the quadratic equation?
5. How would the graph of the equation change if \(b\) or \(c\) were different?
6. Can this quadratic equation have real roots? How can you determine them?
7. How do you solve this quadratic equation using factoring?
8. What are the applications of quadratic equations in real-world scenarios?

**Tip:** When solving systems of equations, check your work by substituting the solutions back into the original equations to ensure they satisfy all conditions.

Learn how to derive a quadratic equation y = ax^2 + bx + c that passes through points (0, 325), (50, 300), and (75, 250). This detailed solution covers the process of solving systems of equations and applying the quadratic formula. Suitable for high school students studying algebra.

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