CONVERTING FROM STANDARD FORM TO VERTEX FORM

Use the information provided to write the vertex form equation of each parabola.

Problem 1 :

y = x² - 4x + 5

Solution :

Step 1 :

The coefficient of x² is 1.

Step 2 :

Subtract 5 on both sides.

y - 5 = x² - 4x + 5 - 5

y - 5 = x² - 4x

Step 3 :

Half of the coefficient of x is 2

Square of half of the coefficient of x is 2² = 4

y - 5 + 4 = x² - 4x + 4

y - 5 + 4 = x² - 2 ⋅ x ⋅ 2 + 2²

y - 1 = (x - 2)²

Step 4 :

By comparing the above equation with vertex form

(y - k) = a(x - h)²

We can get the vertex.

(h, k) = (2, 1)

Since the coefficient of x² is positive the parabola opens upward.

Use the information provided to write the vertex form equation of each parabola.

Problem 2 :

y = x² - 16x + 70

Solution :

Step 1 :

The coefficient of x² is 1.

Step 2 :

Subtract 70 on both sides.

y - 70 = x² - 16x + 70 - 70

y - 70 = x² - 16x

Step 3 :

Half of the coefficient of x is 8

Square of half of the coefficient of x is 8² = 64

y - 70 + 64 = x² - 16x + 64

y - 6 = x² - 2 ⋅ x ⋅ 8 + 64

y - 6 = (x - 8)²

Step 4 :

By comparing the above equation with vertex form

(y - k) = a(x - h)²

We can get the vertex.

(h, k) = (8, 6)

Since the coefficient of x² is positive the parabola opens upward.

Problem 3 :

y = x² - 4x + 2

Solution :

Step 1 :

The coefficient of x² is 1.

Step 2 :

Subtract 2 on both sides.

y - 2 = x² - 4x + 2 - 2

y - 2 = x² - 4x

Step 3 :

Half of the coefficient of x is 2

Square of half of the coefficient of x is 2² = 4

y - 2 + 4 = x² - 4x + 4

y - 2 + 4 = x² - 2 ⋅ x ⋅ 2 + 2²

y + 2 = (x - 2)²

Step 4 :

By comparing the above equation with vertex form

(y - k) = a(x - h)²

We can get the vertex.

(h, k) = (2, -2)

Since the coefficient of x² is positive the parabola opens upward.

Problem 4 :

y  = -3x² + 48x - 187

Solution :

y  = -3x² + 48x - 187

y  = -3[x² - 16x] - 187

Step 1 :

The coefficient of x² is not 1.

Step 2 :

Adding 187 on both sides.

y + 187 = -3[x² - 16x] - 187 + 187

y + 187 = -3[x² - 16x]    

y + 187 = -3(x² - 2 ⋅ x ⋅ 8 + 8² - 8²)

y + 187 = -3[(x - 8)² - 64]

y + 187 = -3(x - 8)² + 192

y + 187 - 192 = -3(x - 8)²

y - 5 = -3(x - 8)²

(y - k) = a(x - h)²

We can get the vertex.

(h, k) = (8, 5)

Since the coefficient of x² is negative the parabola opens downward.

Problem 5 :

y = -2x² - 12x - 12

Solution :

y  = -2x² - 12x - 12

Step 1 :

The coefficient of x² is not 1. 

Step 2 :

Adding 12 on each sides.

y + 12 = -2x² - 12x

y + 12= -2[x² + 6x]

y + 12 = -2(x² + 2 ⋅ x ⋅ 3 + 3² - 3²)

y + 12 = -2[(x + 3)² - 9]

y + 12 = -2(x + 3)² + 18

y + 12 - 18 = -2(x + 3)²

y - 6 = -2(x + 3)²

(y - k) = a(x - h)²

We can get the vertex.

(h, k) = (-3, 6)

Since the coefficient of x² is negative the parabola opens downward.

Problem 6 :

y = 3x² + 18x + 18

Solution :

y  = 3x² + 18x + 18

Step 1 :

The coefficient of x² is not 1. 

Step 2 :

Subtracting 18 on each sides.

y - 18 = 3x² + 18x

y - 18= 3[x² + 6x]

y - 18 = 3(x² + 2 ⋅ x ⋅ 3 + 3² - 3²)

y - 18 = 3[(x + 3)² - 9]

y - 18 = 3(x + 3)² - 27

y - 18 + 27 = 3(x + 3)²

y + 9 = 3(x + 3)²

(y - k) = a(x - h)²

We can get the vertex.

(h, k) = (-3, -9)

Since the coefficient of x² is positive the parabola opens upward.

Problem 7 :

y = 2x² + 3

Solution :

y = 2x² + 0x + 3

Step 1 :

The coefficient of x² is not 1. 

Step 2 :

Subtracting 3 on each sides.

y - 3 = 2x² + 0x

y - 3= x[2x + 0]

(y - k) = a(x - h)²

We can get the vertex.

(h, k) = (0, 3)

Since the coefficient of x² is positive the parabola opens upward.

Problem 8 :

y = 4x² - 56x + 200

Solution :

y  = 4x² - 56x + 200

Step 1 :

The coefficient of x² is not 1. 

Step 2 :

Subtracting 200 on each sides.

y - 200 = 4x² - 56x

y - 200= 4[x² - 14x]

y - 200 = 4(x² - 2 ⋅ x ⋅ 7 + 7² - 7²)

y - 200= 4[(x - 7)² - 49]

y - 200 = 4(x - 7)² - 196

y - 200 + 196 = 4(x - 7)²

y - 4 = 4(x - 7)²

(y - k) = a(x - h)²

We can get the vertex.

(h, k) = (7, 4)

Since the coefficient of x² is positive the parabola opens upward.