*y*=*ax*2+*bx*+*c*  using (75,250) (100,225) (150,250)

To find the quadratic equation \( y = ax^2 + bx + c \) that passes through the points \((75, 250)\), \((100, 225)\), and \((150, 250)\), 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 \((75, 250)\):
   \[
   250 = a(75)^2 + b(75) + c
   \]
   Simplifying:
   \[
   250 = 5625a + 75b + c \quad \text{(Equation 1)}
   \]

2. For \((100, 225)\):
   \[
   225 = a(100)^2 + b(100) + c
   \]
   Simplifying:
   \[
   225 = 10000a + 100b + c \quad \text{(Equation 2)}
   \]

3. For \((150, 250)\):
   \[
   250 = a(150)^2 + b(150) + c
   \]
   Simplifying:
   \[
   250 = 22500a + 150b + c \quad \text{(Equation 3)}
   \]

### Step 2: Form the system of linear equations

Now, we have the following system of equations:

\[
5625a + 75b + c = 250 \quad \text{(Equation 1)}
\]
\[
10000a + 100b + c = 225 \quad \text{(Equation 2)}
\]
\[
22500a + 150b + c = 250 \quad \text{(Equation 3)}
\]

### Step 3: Solve the system of equations

#### Step 3.1: Eliminate \(c\) by subtracting Equation 1 from Equation 2 and Equation 2 from Equation 3.

Subtracting Equation 1 from Equation 2:

\[
(10000a + 100b + c) - (5625a + 75b + c) = 225 - 250
\]
Simplifying:
\[
4375a + 25b = -25 \quad \text{(Equation 4)}
\]
Dividing by 25:
\[
175a + b = -1 \quad \text{(Equation 4 simplified)}
\]

Subtracting Equation 2 from Equation 3:

\[
(22500a + 150b + c) - (10000a + 100b + c) = 250 - 225
\]
Simplifying:
\[
12500a + 50b = 25 \quad \text{(Equation 5)}
\]
Dividing by 25:
\[
500a + 2b = 1 \quad \text{(Equation 5 simplified)}
\]

#### Step 3.2: Solve the system of linear equations obtained from Equation 4 and Equation 5.

Multiply Equation 4 by 2:
\[
350a + 2b = -2
\]
Now subtract this from Equation 5:
\[
(500a + 2b) - (350a + 2b) = 1 - (-2)
\]
Simplifying:
\[
150a = 3 \implies a = \frac{1}{50}
\]

Substitute \(a = \frac{1}{50}\) into Equation 4:
\[
175\left(\frac{1}{50}\right) + b = -1
\]
Simplifying:
\[
\frac{175}{50} + b = -1 \implies \frac{7}{2} + b = -1 \implies b = -\frac{9}{2}
\]

### Step 4: Find \(c\)

Substitute \(a = \frac{1}{50}\) and \(b = -\frac{9}{2}\) into Equation 1:

\[
5625\left(\frac{1}{50}\right) + 75\left(-\frac{9}{2}\right) + c = 250
\]
Simplifying:
\[
112.5 - 337.5 + c = 250 \implies c = 250 + 225 \implies c = 475
\]

### Step 5: Write the final equation

The quadratic equation is:

\[
y = \frac{1}{50}x^2 - \frac{9}{2}x + 475
\]

### Summary

The quadratic equation that passes through the points \((75, 250)\), \((100, 225)\), and \((150, 250)\) is:

\[
y = \frac{1}{50}x^2 - \frac{9}{2}x + 475
\]

Would you like further details or have any questions?

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**Related Questions:**

1. How does the vertex of this parabola compare to the points used?
2. How can you convert this quadratic equation to vertex form?
3. What is the axis of symmetry for the quadratic equation?
4. Can this quadratic equation have real roots, and how would you find them?
5. What effect does changing the coefficient \(a\) have on the parabola's shape?
6. How does the discriminant relate to the nature of the roots of this quadratic equation?
7. How can you solve this quadratic equation using the quadratic formula?
8. What are some practical applications of finding a quadratic equation from points?

**Tip:** When working with quadratic equations, always check your solutions by plugging them back into the original equations to ensure they satisfy all given conditions.

Learn how to find the quadratic equation y = ax^2 + bx + c passing through points (75, 250), (100, 225), and (150, 250). Step-by-step solution provided. Suitable for advanced high school students.

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