Table of Contents

Last modified on August 3rd, 2023

A vertex form is an alternative form of writing the quadratic equation, usually written in the standard form as ax^{2} + 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) ^{2}Â – 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)

Let us consider the quadratic equation y = x^{2} + 12x + 32, to show how a quadratic equation in standard form is converted to its vertex form.

**Step 1**: Isolating the x^{2} and x terms to one side of the equation

y = x^{2} + 12x + 32

=> y – 32 = x^{2} + 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 = x^{2} + 12x + 36

=> y + 4 = x^{2} + 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 = x^{2} + 12x + 32 is y = (x + 6)^{2} – 4

**Write the vertex form of the quadratic equation y = x ^{2} + 8x + 16. Also, find the vertex of the parabola so formed.**

Solution:

Isolating the x^{2} and x terms to one side of the equation

y = x^{2} + 8x + 16

=> y -16 = x^{2}Â + 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 = x^{2}Â + 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 = x^{2} + 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 = ah^{2} + bh + c. Substituting the value of ‘h’ from the above step

Let us consider the above equation y = x^{2} + 8x + 16

Comparing this equation with the standard form of the equation ax^{2} + 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 = ah^{2} + 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 = x^{2} + 8x + 16is y = (x + 4)^{2}

**Write the vertex form of the quadratic equation y = 2x ^{2} + 12x + 13.**

Solution:

Comparing this equation with the standard form of the equation ax^{2} + 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 = ah^{2}Â + 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

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 ax^{2} + bx + c = 0 form.

The steps to follow are:

**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 = x^{2} â€“ 2 Ã— x Ã— 3 + (3)^{2}

= x^{2} â€“ 6x + 9

Multiplying with 2 using the distributive property

= 2(x^{2} â€“ 6x + 9)

= 2x^{2} â€“ 12x + 18

Combining the like terms

= 2x^{2} â€“ 12x + 18 + 4

= 2x^{2} â€“ 12x + 22

Thus, the standard form of the equation y = 2(x – 3)^{2} + 4 is2x^{2} â€“ 12x + 22

Last modified on August 3rd, 2023