How To Find Parabola With Vertex And Point

How to Find the Equation of a Parabola Given Its Vertex and a Point

Understanding how to construct the equation of a parabola when you know its vertex and one other point is a fundamental skill in algebra and analytic geometry. This knowledge transforms abstract concepts into concrete equations, allowing you to model everything from the arc of a thrown ball to the shape of a satellite dish. The process hinges on the vertex form of a quadratic equation, a powerful tool that places the parabola’s most critical feature—its turning point—at the forefront. By systematically substituting your known values, you can unlock the complete equation, revealing the parabola’s precise width and direction.

Understanding the Vertex Form: The Key to the Solution

The standard form of a quadratic equation is y = ax² + bx + c. While useful, this form hides the location of the vertex. The vertex form, y = a(x - h)² + k, is specifically designed for our purpose. Here, the ordered pair (h, k) is the vertex of the parabola. The coefficient a controls the parabola’s width and direction:

• If a > 0, the parabola opens upwards (like a smile).
• If a < 0, the parabola opens downwards (like a frown).
• The absolute value of a determines the width: |a| > 1 makes it narrower; 0 < |a| < 1 makes it wider.

This form is our blueprint. We are given (h, k) and one other point (x₁, y₁). Our single, crucial task is to solve for the value of a. Once a, h, and k are known, the equation is complete.

Step-by-Step Method 1: Direct Substitution into Vertex Form

This is the most straightforward and commonly used approach.

Step 1: Write down the vertex form. Start with the generic template: y = a(x - h)² + k

Step 2: Substitute the vertex coordinates. Plug the given vertex (h, k) directly into the equation. This eliminates h and k, leaving a as the only unknown. Example: Vertex is (3, -2). y = a(x - 3)² + (-2) which simplifies to y = a(x - 3)² - 2.

Step 3: Substitute the coordinates of the given point. Take the second point (x₁, y₁) and substitute its x and y values into the equation from Step 2. This creates an equation with only one variable: a. Example: The point is (5, 6). 6 = a(5 - 3)² - 2

Step 4: Solve for the coefficient a. Simplify the equation and isolate a.

1. Simplify inside the parentheses: (5 - 3) = 2
2. Square the result: 2² = 4
3. The equation becomes: 6 = a * 4 - 2
4. Add 2 to both sides: 8 = 4a
5. Divide by 4: a = 2

Step 5: Write the final equation. Substitute the found value of a back into the equation from Step 2. y = 2(x - 3)² - 2 This is the complete equation of the parabola with vertex (3, -2) passing through (5, 6).

Step-by-Step Method 2: Using the Standard Form (Alternative Approach)

Sometimes, you may need the equation in standard form (y = ax² + bx + c). You can still start with the vertex form to find a, then expand.

Step 1 & 2: Follow Steps 1 and 2 from Method 1 to get an equation with a as the only unknown. y = a(x - h)² + k

Step 3: Substitute the given point (x₁, y₁) and solve for a, as in Method 1.

Step 4: Expand the vertex form. Once a is known, expand (x - h)² to x² - 2hx + h². Then distribute a and combine with k. Continuing the example: y = 2(x - 3)² - 2

1. Expand: (x - 3)² = x² - 6x + 9
2. Distribute a=2: 2(x² - 6x + 9) = 2x² - 12x + 18
3. Add k (-2): y = 2x² - 12x + 18 - 2
4. Combine constants: y = 2x² - 12x + 16

Step 5: Verify. You can plug the original point (5, 6) into this standard form to check: 2*(5)² - 12*5 + 16 = 2*25 - 60 + 16 = 50 - 60 + 16 = 6. It checks out.

The Science Behind the Method: Why Vertex Form Works

The vertex form is not an arbitrary formula; it is derived directly from the process of completing the square applied to the standard form. Starting with y = ax² + bx + c, completing the square on the x terms allows you to rewrite it as y = a(x - h)² + k, where h = -b/(2a) and k = c - b²/(4a).