Standard Form to Vertex Form

A quadratic equation is a type of polynomial equation of degree two, which means the highest exponent of the variable is two. Quadratic equations are used to model various real-world situations, such as projectile motion, area problems, and optimization scenarios.

The solutions to a quadratic equation, known as the roots, can be found using various methods, including factoring, completing the square, and the quadratic formula.

Standard Form

The standard form of a quadratic equation is typically written as

ax² + bx + c = 0

Where:

• x is the variable,
• a, b, and c are constants with a ≠ 0.

Examples of Standard Form

• 2x − 4x + 1 = 0; in this equation, a = 2, b = −4, and c = 1.
• 3x + 5x − 2 = 0; in this equation, a = 3, b = 5, and c = -2.
• x² - 4x + 4 = 0; in this equation, a = 1, b = -4, and c = 4.

Vertex Form

The vertex form of a quadratic equation is a way to express the equation such that it highlights the vertex.

y = a(x − h)² + k

Where:

• (h, k) is the vertex.
• "a" is the same coefficient as in the standard form ax + bx + c and affects the width and direction of the graph.

Examples of Vertex Form

• y = 2(x − 1) + 3; in this equation a = 2, h = 1, k = 3, and the vertex is (1, 3).
• y = -3(x + 4) - 5; in this equation a = -3, h = -4, k = -5, and the vertex is (-4, -5).
• y = (1/2)(x − 2) + 7; in this equation a = 1/2, h = 2, k = 7, and the vertex is (2, 7).

Conversion from Standard Form to Vertex Form

Converting a quadratic equation from standard form ax + bx + c to vertex form a(x − h) + k involves completing the square. Here's a step-by-step guide to the conversion process:

Step 1: Start with the standard form.

y = ax² + bx + c

Step 2: Factor out the coefficient of x² from the x-terms:

If a ≠ 1, factor out a from the x² and x terms: y = a[x + (b/a)x] + c

Step 3: Complete the square:

Take the coefficient of x, divide it by 2, and square it: .

Add and subtract this square inside the parentheses:

Simplify inside the parentheses to form a perfect square trinomial:

Step 4: Simplify the equation:

Distribute a and simplify the constant terms:

⇒

Step 4: Combine the constants:

Combine the constants to get the final vertex form:

In the vertex form , the vertex (h, k) can be identified as

Example: Convert y = 2x² + 8x + 5 to vertex form.

Solution:

Step 1: Start with the standard form:

y = 2x² + 8x + 5

Step 2: Factor out the coefficient of x² from the x-terms:

y = 2(x² + 4x) + 5

Step 3: Complete the square:

Take the coefficient of x, divide it by 2, and square it: (4/2)² = 4.

Add and subtract this square inside the parentheses:

y = 2 [x² + 4x + 4 - 4] + 5

⇒ y = 2[(x + 2)² - 4] + 5

Step 4: Simplify the equation:

Distribute 2 and simplify the constants:

y = 2(x + 2)² - 8 + 5

⇒ y = 2(x + 2)² - 3

So, the vertex form of y = 2x² + 8x + 5 is:

y = 2(x + 2)² - 3

The vertex is (-2, -3).

Related Articles:

• Polynomials
• How to find the Discriminant of a Quadratic Equation
• Real-Life Applications of Quadratic Equations
• Standard Equation of a Parabola
• Solving Quadratic Equations by Factoring
• Graphing Quadratic Functions 

Solved Examples of Converting Standard Form to Vertex Form

Example 1: Convert the quadratic equation y = -3x²² + 6x - 1 from standard form to vertex form.

Solution:

Given: y = -3x² + 6x - 1

Factor out the coefficient of x² from the first two terms

y = -3(x² - 2x) - 1

Take the coefficient of x, which is -2, divide it by 2, and square it: (-2/2)²

Add and subtract this square inside the parentheses:

y = -3(x² - 2x + 1 - 1) - 1

⇒ y = -3((x - 1)² - 1) - 1

⇒ y = -3(x - 1)² + 3 - 1

⇒ y = -3(x - 1)² + 2

So, the vertex form of y = -3x² + 6x - 1 is:

y = -3(x - 1)² + 2

Example 2: Convert the quadratic equation y = x²² - 4x + 7 from standard form to vertex form.

Solution:

Given: y = x² - 4x + 7

Factor out the coefficient of x² from the first two terms (coefficient is 1 here, so no factoring needed)

y = x² - 4x + 7

Take the coefficient of x, which is -4, divide it by 2, and square it: (-4/2)² = 4

Add and subtract this square inside the parentheses:

y = x² - 4x + 4 - 4 + 7

⇒ y = (x - 2)² - 4 + 7

⇒ y = (x - 2)² + 3

So, the vertex form of y = x² - 4x + 7 is:

y = (x - 2)² + 3

The vertex is (2, 3).

Example 3: Convert the quadratic equation y = 2x²² - 8x - 5 from standard form to vertex form.

Solution:

Given: y = 2x² - 8x - 5

Factor out the coefficient of x² from the first two terms

y = 2(x² - 4x) - 5

Take the coefficient of x, which is 4, divide it by 2, and square it: (4/2)² = 4

Add and subtract this square inside the parentheses:

y = 2(x² + 4x + 4 - 4) - 5

y = 2((x - 2)² - 4) - 5

Distribute the 2 and combine like terms:

y = 2(x - 2)² - 8 - 5

y = 2(x - 2)² - 13

So, the vertex form of y = 2x² + 8x + 5 is:

y = 2(x + 2)² - 3

Practice Problems on Standard Form to Vertex Form

Problem 1: Convert the quadratic equation y = 3x² + 12x + 5 from standard form to vertex form.

Problem 2: Convert the quadratic equation y = -2x² + 8x - 6 from standard form to vertex form.

Problem 3: Convert the quadratic equation y = x² - 6x + 10 from standard form to vertex form.

Problem 4: Convert the quadratic equation y = -4x² + 16x - 7 from standard form to vertex form.

Problem 5: Convert the quadratic equation y = 5x² + 20x + 15 from standard form to vertex form.