HOW TO FIND THE EQUATION OF A CUBIC FUNCTION FROM GRAPH IN VERTEX FORM

Write a cubic function in the form 

h(x) = a (x - h)³ + k

Problem 1 :

Solution :

By comparing the given graph with y = x³, we notice that the curve is moved

• horizontally 2 units to the right 
• vertically 1 unit up.

So, h = 2 and k = 1

h(x) = a (x - 2)³ + 1

y = a (x - 2)³ + 1 ------(1)

The curve passes through the point (1, -2) and (3, 4)

Applying the point (1, -2)

-2 = a(1 - 2)³ + 1

-2 = a(-1)³ + 1

-2 = -a + 1

-a = -2 - 1

a = 3

Applying the value of a in (1), we get

y = 3 (x - 2)³ + 1

Problem 2 :

Solution :

By comparing the given graph with y = x³, we notice that the curve is moved

• horizontally 2 units to the left
• vertically 2 units down.

So, h = -2 and k = -2

h(x) = a (x - (- 2))³ + (-2)

h(x) = a (x + 2)³ - 2 

y = a (x + 2)³ - 2 ------(1)

The curve passes through the point (-1, 1) and (-3, -5)

Applying the point (-1, 1)

1 = a(-1 + 2)³ - 2

1 + 2 = a(1)

a = 3

Applying the value of a in (1), we get

y = 3 (x + 2)³ - 2

Problem 3 :

Solution :

By comparing the given graph with y = x³, we notice that the curve is moved

• There is no horizontal movements.
• vertically 1 unit up.

So, h = 0 and k = 1

h(x) = a (x - 0)³ + 1

h(x) = a x³ + 1 

y = a x³ + 1  ------(1)

The curve passes through the point (-1, 2) and (0, -1)

Applying the point (-1, 2)

2 = a(-1)³ + 1

2 = -a + 1

2 - 1 = -a

-a = 1

a = -1

Applying the value of a in (1), we get

y = -x³ + 1

Problem 4 :

Solution :

By comparing the given graph with y = x³, we notice that the curve is moved as follows.

• Horizontally moved 1 unit left.
• No vertical movements.

So, h = -1 and k = 0

h(x) = a (x - (-1))³ + 0

h(x) = a (x + 1)³

y = a (x + 1)³  ------(1)

The curve passes through the point (-2, 1) and (0, -1)

Applying the point (-2, 1)

1 = a (-2 + 1)³

1 = -a

a = -1

Applying the value of a in (1), we get

y =  -1 (x - 1)³