Find the quadratic function of a graph that goes through (0,15), (3,0), (5,0)

To find the quadratic function that goes through the points \((0, 15)\), \((3, 0)\), and \((5, 0)\), we will use the general form of a quadratic equation:

\[
y = ax^2 + bx + c
\]

We are given the three points, which will allow us to create a system of equations.

1. **For the point (0, 15)**: When \(x = 0\), \(y = 15\):
\[
15 = a(0)^2 + b(0) + c \quad \Rightarrow \quad c = 15
\]

2. **For the point (3, 0)**: When \(x = 3\), \(y = 0\):
\[
0 = a(3)^2 + b(3) + 15 \quad \Rightarrow \quad 0 = 9a + 3b + 15
\]
This simplifies to:
\[
9a + 3b = -15
\quad \text{(Equation 1)}
\]

3. **For the point (5, 0)**: When \(x = 5\), \(y = 0\):
\[
0 = a(5)^2 + b(5) + 15 \quad \Rightarrow \quad 0 = 25a + 5b + 15
\]
This simplifies to:
\[
25a + 5b = -15
\quad \text{(Equation 2)}
\]

Now, we solve this system of equations:

### Step 1: Solve for \(b\) in terms of \(a\)
From Equation 1:
\[
9a + 3b = -15 \quad \Rightarrow \quad 3b = -15 - 9a \quad \Rightarrow \quad b = -5 - 3a
\]

### Step 2: Substitute \(b = -5 - 3a\) into Equation 2
\[
25a + 5(-5 - 3a) = -15
\]
Simplify:
\[
25a - 25 - 15a = -15 \quad \Rightarrow \quad 10a - 25 = -15
\]
Solve for \(a\):
\[
10a = 10 \quad \Rightarrow \quad a = 1
\]

### Step 3: Solve for \(b\)
Substitute \(a = 1\) into \(b = -5 - 3a\):
\[
b = -5 - 3(1) = -8
\]

### Step 4: Write the quadratic function
Now that we have \(a = 1\), \(b = -8\), and \(c = 15\), the quadratic function is:

\[
y = x^2 - 8x + 15
\]

This is the quadratic function

Find the quadratic function of a graph that goes through (0,15), (3,0), (5,0).

Learn how to determine the quadratic function that passes through the points (0,15), (3,0), and (5,0). This problem involves solving a system of equations to find the coefficients of the quadratic equation. Suitable for students in Grades 9-11.

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