solve using quadratic formula y=ax^2+bx+c (0,100), (40,80), (60,100)

To solve for the quadratic equation \( y = ax^2 + bx + c \) using the given points \((0, 100)\), \((40, 80)\), and \((60, 100)\), we need to set up a system of equations. Each point provides a relationship between \( x \) and \( y \), which we can use to find the coefficients \( a \), \( b \), and \( c \).

### Step 1: Set Up the Equations
Plug in the points into the quadratic equation \( y = ax^2 + bx + c \):

1. For \((0, 100)\):
   \[
   100 = a(0)^2 + b(0) + c
   \]
   \[
   c = 100
   \]

2. For \((40, 80)\):
   \[
   80 = a(40)^2 + b(40) + c
   \]
   \[
   80 = 1600a + 40b + 100
   \]
   Simplifying:
   \[
   1600a + 40b = -20 \quad \text{(Equation 1)}
   \]

3. For \((60, 100)\):
   \[
   100 = a(60)^2 + b(60) + c
   \]
   \[
   100 = 3600a + 60b + 100
   \]
   Simplifying:
   \[
   3600a + 60b = 0 \quad \text{(Equation 2)}
   \]

### Step 2: Solve the System of Equations
We have the following system of linear equations:
\[
1600a + 40b = -20 \quad \text{(Equation 1)}
\]
\[
3600a + 60b = 0 \quad \text{(Equation 2)}
\]

#### Solve for \( b \) in terms of \( a \):
From Equation 2:
\[
3600a + 60b = 0
\]
\[
60b = -3600a
\]
\[
b = -60a
\]

#### Substitute \( b = -60a \) into Equation 1:
\[
1600a + 40(-60a) = -20
\]
\[
1600a - 2400a = -20
\]
\[
-800a = -20
\]
\[
a = \frac{-20}{-800} = \frac{1}{40}
\]

#### Substitute \( a = \frac{1}{40} \) into \( b = -60a \):
\[
b = -60 \times \frac{1}{40} = -\frac{60}{40} = -\frac{3}{2}
\]

### Final Equation
Now that we have \( a = \frac{1}{40} \), \( b = -\frac{3}{2} \), and \( c = 100 \), the quadratic equation is:
\[
y = \frac{1}{40}x^2 - \frac{3}{2}x + 100
\]

### Summary
The quadratic equation that passes through the points \((0, 100)\), \((40, 80)\), and \((60, 100)\) is:
\[
\boxed{y = \frac{1}{40}x^2 - \frac{3}{2}x + 100}
\]

Would you like to see more details, or do you have any further questions?

### Related Questions
1. How can we determine the vertex of the quadratic function?
2. How do we calculate the discriminant for this quadratic equation?
3. What are the x-intercepts (roots) of this quadratic equation?
4. How would the graph of this quadratic function look?
5. What is the significance of the coefficient \( c \) in the equation?

### Tip
When solving for coefficients in a quadratic equation, always ensure that you systematically substitute and solve the system of equations to avoid mistakes.

Learn how to solve the quadratic equation y = ax^2 + bx + c using points (0, 100), (40, 80), and (60, 100). Detailed step-by-step solution covering system of equations and quadratic formula. Suitable for high school students in Grades 10-12.

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