# 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  = -3x+ 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.

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