To find the equation of a parabola that passes through three points, we assume the general form of a quadratic equation:

$y = ax^2 + bx + c$

Given that the parabola passes through the points $(0, -2)$, $(2, -6)$, and $(5, 3)$, we can substitute these points into the equation to form a system of three equations.

Step 1: Set up the system of equations

1. Using the point $(0, -2)$: $-2 = a(0)^2 + b(0) + c \quad \Rightarrow \quad c = -2$

2. Using the point $(2, -6)$: $-6 = a(2)^2 + b(2) + c \quad \Rightarrow \quad -6 = 4a + 2b - 2 \quad \Rightarrow \quad 4a + 2b = -4$

3. Using the point $(5, 3)$: $3 = a(5)^2 + b(5) + c \quad \Rightarrow \quad 3 = 25a + 5b - 2 \quad \Rightarrow \quad 25a + 5b = 5$

Step 2: Solve the system of equations

We now have two equations to solve for $a$ and $b$:

1. $4a + 2b = -4$
2. $25a + 5b = 5$

We can simplify and solve this system using substitution or elimination.

Elimination method:

First, multiply the first equation by 5 and the second equation by 2 to eliminate $b$:

$5(4a + 2b) = 5(-4) \quad \Rightarrow \quad 20a + 10b = -20$ $2(25a + 5b) = 2(5) \quad \Rightarrow \quad 50a + 10b = 10$

Now subtract the first equation from the second:

$(50a + 10b) - (20a + 10b) = 10 - (-20) \quad \Rightarrow \quad 30a = 30 \quad \Rightarrow \quad a = 1$

Solve for $b$:

Substitute $a = 1$ into $4a + 2b = -4$:

$4(1) + 2b = -4 \quad \Rightarrow \quad 4 + 2b = -4 \quad \Rightarrow \quad 2b = -8 \quad \Rightarrow \quad b = -4$

Step 3: Write the equation

Now that we have $a = 1$, $b = -4$, and $c = -2$, the equation of the parabola is:

$y = x^2 - 4x - 2$

Final Answer:

The equation of the parabola is: $y = x^2 - 4x - 2$

Would you like more details or have any questions?

Here are five related questions to expand on this problem:

1. How would the parabola change if the vertex form of the equation was needed?
2. How can we graph the parabola $y = x^2 - 4x - 2$ using its vertex and intercepts?
3. How can we find the focus and directrix of the parabola based on its equation?
4. What is the axis of symmetry of this parabola?
5. How does changing the coefficient of $x^2$ affect the shape of the parabola?

Tip: The vertex of a parabola in standard form $y = ax^2 + bx + c$ can be found using the formula $x = -\frac{b}{2a}$.