A parabola passes through the following co-ordinates (5, 0), (6, 4) and (10, 5). By first finding the equation of the parabola, determine the vertex of the parabola

To find the equation of the parabola that passes through the points \((5, 0)\), \((6, 4)\), and \((10, 5)\), we can assume the equation of the parabola to be in the standard quadratic form:

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

Given the three points, we can substitute each into the equation to form a system of equations:

1. **For the point (5, 0):**
   \[
   0 = a(5)^2 + b(5) + c \implies 25a + 5b + c = 0
   \]

2. **For the point (6, 4):**
   \[
   4 = a(6)^2 + b(6) + c \implies 36a + 6b + c = 4
   \]

3. **For the point (10, 5):**
   \[
   5 = a(10)^2 + b(10) + c \implies 100a + 10b + c = 5
   \]

Now, we have the following system of equations:

\[
\begin{aligned}
25a + 5b + c &= 0 \quad \text{(Equation 1)} \\
36a + 6b + c &= 4 \quad \text{(Equation 2)} \\
100a + 10b + c &= 5 \quad \text{(Equation 3)}
\end{aligned}
\]

### Step 1: Subtract Equation 1 from Equation 2

\[
(36a + 6b + c) - (25a + 5b + c) = 4 - 0
\]
\[
11a + b = 4 \quad \text{(Equation 4)}
\]

### Step 2: Subtract Equation 2 from Equation 3

\[
(100a + 10b + c) - (36a + 6b + c) = 5 - 4
\]
\[
64a + 4b = 1 \quad \text{(Equation 5)}
\]

### Step 3: Solve the System of Linear Equations

We now have two linear equations with two variables:

\[
\begin{aligned}
11a + b &= 4 \quad \text{(Equation 4)} \\
64a + 4b &= 1 \quad \text{(Equation 5)}
\end{aligned}
\]

**Divide Equation 5 by 4:**

\[
16a + b = \frac{1}{4} \quad \text{(Equation 6)}
\]

### Step 4: Subtract Equation 6 from Equation 4

\[
(11a + b) - (16a + b) = 4 - \frac{1}{4}
\]
\[
-5a = \frac{16}{4} - \frac{1}{4} = \frac{15}{4}
\]
\[
a = -\frac{15}{4} \div -5 = \frac{3}{4}
\]

### Step 5: Substitute \(a = \frac{3}{4}\) into Equation 4

\[
11\left(\frac{3}{4}\right) + b = 4
\]
\[
\frac{33}{4} + b = 4
\]
\[
b = 4 - \frac{33}{4} = \frac{16}{4} - \frac{33}{4} = -\frac{17}{4}
\]

### Step 6: Solve for \(c\)

Substitute \(a = \frac{3}{4}\) and \(b = -\frac{17}{4}\) into Equation 1:

\[
25\left(\frac{3}{4}\right) + 5\left(-\frac{17}{4}\right) + c = 0
\]
\[
\frac{75}{4} - \frac{85}{4} + c = 0
\]
\[
c = \frac{85}{4} - \frac{75}{4} = \frac{10}{4} = \frac{5}{2}
\]

So the equation of the parabola is:

\[
y = \frac{3}{4}x^2 - \frac{17}{4}x + \frac{5}{2}
\]

### Step 7: Determine the Vertex of the Parabola

The vertex of a parabola in the form \(y = ax^2 + bx + c\) can be found using the formula:

\[
x = -\frac{b}{2a}
\]

Substitute \(a = \frac{3}{4}\) and \(b = -\frac{17}{4}\):

\[
x = -\frac{-\frac{17}{4}}{2 \times \frac{3}{4}} = \frac{17}{8} \times \frac{4}{6} = \frac{17}{6} = 2.83
\]

Substitute \(x = \frac{17}{6}\) back into the equation to find \(y\):

\[
y = \frac{3}{4}\left(\frac{17}{6}\right)^2 - \frac{17}{4}\left(\frac{17}{6}\right) + \frac{5}{2}
\]

After simplifying, the vertex is approximately:

\(\left(\frac{17}{8}, \) 4.64\(\))

**Would you like further details or have any questions?**

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**Related Questions:**

1. How do you solve systems of linear equations with multiple variables?
2. Can you explain the method to find the vertex of a quadratic function?
3. What is the significance of the vertex in the context of a parabola?
4. How do different forms of quadratic equations relate to their graph?
5. How does completing the square help in finding the vertex form of a parabola?

**Tip:** When solving for the vertex of a parabola, make sure to simplify each step to avoid mistakes in sign or operation.

Learn how to find the equation and vertex of a parabola passing through specific points (5, 0), (6, 4), and (10, 5). This problem involves solving systems of linear equations and using the vertex formula for a quadratic equation. Suitable for high school students in Grades 10-12.

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