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Last modified on November 9th, 2022

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 ≠ 0 and

- (h, k) is the vertex of the parabola
- ‘a’ is the leading constant (constant) that tells us whether the parabola is facing upwards or downwards. If ‘a’ is positive, the parabola is facing upwards, and if ‘a’ is negative, the parabola is facing downwards
- ‘h’ represents horizontal translation (shift left or right from x = 0)
- ‘k’ 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
- ‘a’ 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(x + 4) ^{2} – 6**

Solution:

Comparing the given equation with the vertex form of the quadratic equation y = a(x – h)^{2} + k, the vertex is at:

h = -4, k = -6

Thus, the vertex of the parabola with equation 3(x + 4)2 – 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 – 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} + 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} + 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 ‘h’ and ‘k’ 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 – (-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} + 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 ‘a’ 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}

= x^{2} – 2 × x × 3 + (3)^{2}

= x^{2} – 6x + 9

Multiplying with 2 using the distributive property

= 2(x^{2} – 6x + 9)

= 2x^{2} – 12x + 18

Combining the like terms

= 2x^{2} – 12x + 18 + 4

= 2x^{2} – 12x + 22

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

Last modified on November 9th, 2022