# Vertex Form of a Quadratic Equation

A vertex form is an alternative form of writing the quadratic equation, usually written in the standard form as ax2 + bx + c = 0. Graphing a quadratic function gives a parabola, which helps find the two roots of the equation.

However, if we need to find the vertex of a parabola, the standard form of the equation is less useful. In such cases, we need to convert the equation into an alternative vertex form, which is written as:

y = a(x – h)2 + k, here a â‰  0 and

• (h, k) is the vertex of the parabola
• â€˜aâ€™ is the leading constant (constant) that tells us whether the parabola is facing upwards or downwards. If â€˜aâ€™ is positive, the parabola is facing upwards, and if â€˜aâ€™ is negative, the parabola is facing downwards
• â€˜hâ€™ represents horizontal translation (shift left or right from x = 0)
• â€˜kâ€™ represents vertical translation (shift up or down from y = 0)

For example, in the equation

y = 4(x – 1)2 + 3

• (1, 3) is the vertex of the parabola
• â€˜aâ€™ is 4, and thus the parabola is facing upwards
• The horizontal translation is 1 unit to the right of x = 0
• The vertical translation is 3 units up from x = 0

Find the vertex of the parabola with equation 3(xÂ + 4)2Â – 6

Solution:

Comparing the given equation with the vertex form of the quadratic equation y = a(x – h)2 + k,Â  the vertex is at:
h = -4, k = -6
Thus, the vertex of the parabola with equation 3(x + 4)2 â€“ 6 is (-4, -6)

## Converting Standard Form to Vertex Form

### By Completing the Square

Let us consider the quadratic equation y = x2 + 12x + 32, to show how a quadratic equation in standard form is converted to its vertex form.

Step 1: Isolating the x2 and x terms to one side of the equation

y = x2 + 12x + 32

=> y – 32 = x2 + 12x

Step 2: Add ${\left(\dfrac{b}{2}\right) ^{2}}$ to both sides of the equation

Here, b = ${\left( \dfrac{12}{2}\right) ^{2}}$ = 36

Thus,

y â€“ 32 + 36 = x2 + 12x + 36

=> y + 4 = x2 + 12x + 36

Step 3: Factor the right side of the equation into a perfect square

=> y + 4 = (x + 6)2

Step 4: Rewriting the equation in terms of y

=> y = (x + 6)2 – 4

Thus the vertex form of the equation y = x2 + 12x + 32 is y = (x + 6)2 – 4

Write the vertex form of the quadratic equation y = x2 + 8x + 16. Also, find the vertex of the parabola so formed.

Solution:

Isolating the x2 and x terms to one side of the equation
y = x2 + 8x + 16
=> y -16 = x2Â + 8x
Adding ${\left(\dfrac{b}{2}\right) ^{2}}$ to both sides of the equation
Here, b = ${\left( \dfrac{8}{2}\right) ^{2}}$ = 16
Thus,
=> y -16 + 16 = x2Â + 8x + 16

Step 3: Factoring the right side of the equation into a perfect square

=> y = (x + 4)2

Thus the vertex form of the equation y = x2 + 8x + 16is y = (x + 4)2, and the vertex of the parabola is (-4, 0)

When working with the vertex form of the quadratic equation, the value of â€˜hâ€™ and â€˜kâ€™ can be found as:

${h=-\dfrac{b}{2a}}$

k = f(h) as  (h, k) lies on the given parabola, k = ah2 + bh + c. Substituting the value of ‘h’ from the above step

Let us consider the above equation y = x2 + 8x + 16

Comparing this equation with the standard form of the equation ax2 + bx + c = 0, we get a = 1, b = 8, c = 16

Thus,

${h=-\dfrac{b}{2a}}$ = ${\dfrac{-8}{2\times 1}}$ = -4

Substituting the value of h in the form k = ah2 + bh + c, we get

k = 1(-4)2 + 8(-4) + 16 = 0

Substituting these values (along with a = 1) in the vertex form y = a(x – h)2 + k, we get

=> y = 1(x â€“ (-4))2 + 0

=> y = (x + 4)2

Thus the vertex form of the equation y = x2 + 8x + 16is y = (x + 4)2

Write the vertex form of the quadratic equation y = 2x2 + 12x + 13.

Solution:

Comparing this equation with the standard form of the equation ax2 + bx + c = 0, we get a = 2, b = 12, c = 13
Thus,
h = ${\dfrac{-12}{2\times 2}}$ = -3
Substituting the value of h in the form k = ah2Â + bh + c, we get
k = 2(-3)2 + 12(-3) + 13
= -5
Substituting these values (along with a = 2) in the vertex form y = a(x – h)2 + k, we get
y = 2(x + 3)2 -5
Thus the vertex form of the equation y = 2×2 + 12x + 13 is y = 2(x + 3)2 – 5

## Converting Vertex Form to Standard Form

To convert vertex form of the quadratic equation to the standard form we need to simplify the vertex form y = a(x – h)2 + k algebraically to get to the ax2 + bx + c = 0 form.

Step 1: Expand the square (x – h)2 using the identity (a – b)2

Step 2: Multiply with â€˜aâ€™ using the distributive property

Step 3: Combine the like terms

Convert equation y = 2(x – 3)2 + 4 to the standard form.

Solution:

Expanding the square (x – 3)2 using the identity (a – b)2
Â = x2 â€“ 2 Ã— x Ã— 3 + (3)2
= x2 â€“ 6x + 9
Multiplying with 2 using the distributive property
= 2(x2 â€“ 6x + 9)
= 2x2 â€“ 12x + 18
Combining the like terms
= 2x2 â€“ 12x + 18 + 4
= 2x2 â€“ 12x + 22
Thus, the standard form of the equation y = 2(x – 3)2 + 4 is2x2 â€“ 12x + 22