FIND THE EQUATION OF A TRANSFORMED QUADRATIC FUNCTION FROM A GRAPH

y = a(x - h)² + k

is the equation of parabola which is having the vertex (h, k)

The sign of a will decide if there is any reflection or not.

• The sign of a is positive, the parabola will open up
• The sign of a is negative, the parabola will open down.

The parabolas shown are the result of translating and/or reflecting the parabola y = x². Find the equation of each parabola.

Problem 1 :

Solution :

Since the parabola opens down, there is a reflection.

a = -1

Horizontally the graph of y = -x² is not moved.

Vertically the graph of y = -x² is moved 4 units up.

h = 0 and k = 4

y = -1(x - 0)² + 4

So, the equation from the given graph is 

y = -1x² + 4

Problem 2 :

Solution :

Since the parabola opens down, there is no reflection.

a = 1

Horizontally the graph of y = x² is moved 3 units to the left.

Vertically the graph of y = -x² is not moved.

h = -3 and k = 0

y = 1(x - (-3))² + 0

So, the equation from the given graph is 

y = (x + 3)²

Problem 3 :

Solution :

Since the parabola opens down, there is reflection.

a = -1

Horizontally the graph of y = -x² is moved 3 units to the right.

Vertically the graph of y = -x² is moved 3 units up.

h = 3 and k = 3

y = -1(x - 3)² + 3

So, the equation from the given graph is 

y = -(x - 3)² + 3

Problem 4 :

Solution :

Since the parabola opens down, there is no reflection.

a = 1

Horizontally the graph of y = x² is moved 2 units to the right.

Vertically the graph of y = x² is moved 3 units down.

h = 2 and k = -3

y = 1(x - 2)² - 3

So, the equation from the given graph is 

y = 1(x - 2)² - 3

Problem 5 :

Solution :

Since the parabola opens down, there is no reflection.

a = 1

Horizontally the graph of y = x² is moved 2 units to the right.

Vertically the graph of y = x² is moved 1 unit up.

h = 2 and k = 1

y = 1(x - 2)² + 1

So, the equation from the given graph is 

y = (x - 2)² + 1

Problem 6 :

Solution :

Since the parabola opens down, there is no reflection.

a = 1

Horizontally the graph of y = x² is moved 2 units left.

Vertically the graph of y = x² is moved 1 unit down.

h = -2 and k = -1

y = 1(x - (-2))² - 1

So, the equation from the given graph is 

y = (x + 2)² - 1

Problem 7 :

Describe the transformation of the graph of the parent quadratic function. Then identify the vertex. 

f(x) = 2(x + 3)² + 2

Solution :

f(x) = 2(x + 3)² + 2

f(x) = 2(x - (-3))² + 2

Comparing with f(x) = x²

• Vertical stretch of 2
• Moving the parabola 2 units up
• Moving the parabola 3 units left.
• Vertex is (-3, 2)

Problem 8 :

write a rule for g described by the transformations of the graph of f. Then identify the vertex.

f(x) = x²

vertical stretch by a factor of 4 and a reflection in the x-axis, followed by a translation 2 units up

Solution :

f(x) = x²

f(x) = -4(x + 2)²

So, the required quadratic function is f(x) = -4(x + 2)²

Problem 9 :

Match the function with its graph. Explain your reasoning.

a) g(x) = 2(x − 1)² − 2

b) g(x) = 1/2 ( x + 1)² − 2

c) g(x) = −2(x − 1)² + 2

d) g(x) = 2(x + 1)² + 2

e) g(x) = −2(x + 1)² − 2

f) g(x) = 2(x − 1)² + 2

Solution :

a) g(x) = 2(x − 1)² − 2

• Vertical stretch of 2
• Moving horizontally 1 unit to the right
• Moving vertically 2 units down.

Option C is correct.

b) g(x) = 1/2 (x + 1)² − 2

• Vertical shrink of 1/2
• Moving horizontally 1 unit to the left
• Moving vertically 2 units down.

Option B is correct.

c) g(x) = −2(x − 1)² + 2

• Vertical stretch of 2, opens down.
• Moving horizontally 1 unit to the right
• Moving vertically 2 units up

Option D is correct.

d) g(x) = 2(x + 1)² + 2

• Vertical stretch of 2, opens up.
• Moving horizontally 1 unit to the left
• Moving vertically 2 units up

Option E is correct.

e) g(x) = −2(x + 1)² − 2

• Vertical stretch of 2, opens down
• Moving horizontally 1 unit to the left
• Moving vertically 2 units down

Option F is correct.

f) g(x) = 2(x − 1)² + 2

• Vertical stretch of 2, opens up.
• Moving horizontally 1 unit to the right
• Moving vertically 2 units up

Option A is correct.