IDENTIFYING TRANSFORMATIONS OF QUADRATIC FUNCTIONS

A quadratic function is a function that can be written in the form

f(x) = a(x โˆ’ h)2 + k

where a โ‰  0.

The U-shaped graph of a quadratic function is called a parabola.

The transformation given below can be done for any quadratic function.

i) translation

ii) stretch or shrink

iii)  reflection

Translation

f(x) = a(x โˆ’ h)2 + k

Horizontal translation :

Moving the graph towards left or right.

h > 0, then move the graph right.

h < 0, then move the graph left.

Vertical translation :

Moving the graph towards up or down.

k > 0, then move the graph up.

k < 0, then move the graph down.

translation-of-quadratic-function

Reflection

Reflection across x-axis :

f(x) = x2

-f(x) = -x2

Reflection across y-axis :

f(x) = x2

f(-x) = (-x)2

reflection-of-quadratic-function

Shrinks and Stretches

Horizontal stretches and shrinks :

f(x) = x2

f(ax) = (ax)2

If a > 1 (horizontal shrink)

If 0 < a < 1 (horizontal stretch)

Vertical stretches and shrinks :

f(x) = x2

a f(x) = ax2

If a > 1 (vertical stretch)

If 0 < a < 1 (vertical shrink)

stretches-and-shrinks-of-quad-fun

Problem 1 :

Which correctly identifies the values of the parameters a, h, and k for the function

๐‘“(๐‘ฅ) = โˆ’(๐‘ฅ โˆ’ 1)2 โˆ’ 4

a.๐‘Ž = 1, โ„Ž = โˆ’1, ๐‘˜ = โˆ’4            b. ๐‘Ž = โˆ’1, โ„Ž = 1, ๐‘˜ = โˆ’4

c. ๐‘Ž = โˆ’1, โ„Ž = โˆ’1, ๐‘˜ = โˆ’4      d. ๐‘Ž = โˆ’1, โ„Ž = 1, ๐‘˜ = 4

Solution :

๐‘“(๐‘ฅ) = โˆ’(๐‘ฅ โˆ’ 1)2 โˆ’ 4

By comparing the given function with general form of quadratic function with vertex (h, k), we get

f(x) = a(x - h)2 + k

Here a = -1, h = 1 and k = -4

Problem 2 :

What is the equation of this graph?

identifying-transofrmationq2

a. y = x2 - 4x       b.  -x- 4x       c. -x2 - 4      d. -x2 + 4

Solution :

Quadratic function in vertex form will be 

f(x) = a(x - h)2 + k

By observing the figure, vertex is (-2, 4) and it passes through the point (0, 0).

h = -2 and k = 4

f(x) = a(x - (-2))2 + 4

f(x) = a(x + 2)2 + 4

It passes through (0, 0).

0 = a(0 + 2)2 + 4

4a = -4

a = -1

Applying the value, we get

f(x) = -1(x + 2)2 + 4

Expanding it, we get

f(x) = -1(x2 + 4x + 4) + 4

f(x) = -1x2 - 4x - 4 + 4

f(x) = -1x2 - 4x

Problem 3 :

Which function includes a translation of 2 two units to the right?

a. ๐‘“(๐‘ฅ) = ๐‘ฅ2 + 2                  b. ๐‘“(๐‘ฅ) = (๐‘ฅ โˆ’ 2)2 โˆ’ 3

c. ๐‘“(๐‘ฅ) = (๐‘ฅ + 2)2 โˆ’ 4         d. ๐‘“(๐‘ฅ) = 2๐‘ฅ2

Solution :

Option a :

๐‘“(๐‘ฅ) = ๐‘ฅ2 + 2

K = 2, vertically moving the graph two units up.

Option b :

๐‘“(๐‘ฅ) = (๐‘ฅ โˆ’ 2)2 โˆ’ 3

here h = 2 and k = -3

so, move the graph horizontally 2 units right and move vertically 3 units. down. then option b is correct.

Problem 4 :

Which function includes a translation of 4 units to the left and a vertical compression to the graph of ๐‘“(๐‘ฅ) = ๐‘ฅ2 ?

a. ๐‘“(๐‘ฅ) = 1/3(๐‘ฅ - 4)2                  b. ๐‘“(๐‘ฅ) = 3(๐‘ฅ โˆ’ 4)2

c. ๐‘“(๐‘ฅ) = 1/3(๐‘ฅ + 4)2                  b. ๐‘“(๐‘ฅ) = 3(๐‘ฅ + 4)2

Solution :

๐‘“(๐‘ฅ) = ๐‘ฅ2

Here h = -4 and the value of a should lie between 0 to 1 since it is vertical compression.

In ๐‘“(๐‘ฅ) = 1/3(๐‘ฅ + 4)2 

h = -4 and a = 1/3 (0 < a < 1)

So, option c is correct.

Problem 5 :

List the sequence of steps required to graph the function

๐‘ฆ = โˆ’(๐‘ฅ โˆ’ 3)2 โˆ’ 2

a. horizontal translation 3 units to the right, vertical compression by a factor of 1, vertical translation 2 units down

b. horizontal translation 3 units to the right, reflection in x-axis, vertical translation 2 units down.

c. horizontal translation 3 units to the left, vertical translation 2 units up, reflection in x-axis.

d. horizontal translation 3 units to the left, reflection in x-axis, vertical translation 2 units down

Solution :

๐‘ฆ = โˆ’(๐‘ฅ โˆ’ 3)2 โˆ’ 2

Reflection across x-axis, h = 3 and k = -2

After reflection across x-axis, the graph is moved 3 units right and 2 units down. So, option b is correct.

Problem 6 :

Which function matches the graph?

identifying-transofrmationq6.png

a. ๐‘“(๐‘ฅ) = (๐‘ฅ + 1)2 + 9       b. ๐‘“(๐‘ฅ) = (๐‘ฅ โˆ’ 1)2 + 9

c. ๐‘“(๐‘ฅ) = โˆ’(๐‘ฅ + 1)2 + 9       d. ๐‘“(๐‘ฅ) = โˆ’(๐‘ฅ โˆ’ 1)2 + 9

Solution :

The given graph is opening down, so reflection across x-axis is made. h = -1 and k = 9. So, option c is the answer.

Problem 7 :

Match the description to its equations. Vertical stretch by a factor of 5.

a. ๐‘“(๐‘ฅ) = 5๐‘ฅ2                   b. ๐‘“(๐‘ฅ) = (5๐‘ฅ)2

c. ๐‘“(๐‘ฅ) = 1/5 (๐‘ฅ)2            d. ๐‘“(๐‘ฅ) = ( 1/5 ๐‘ฅ)2

Solution :

Vertical stretch means, the value of a would be greater than 1. So, the answer will be option a.

Problem 8 :

If 0 < ๐‘Ž < 1, what would be the transformation of, ๐‘Ž , from the quadratic parent function, ๐‘“(๐‘ฅ) = ๐‘ฅ2 ?

a. ๐‘ฃ๐‘’๐‘Ÿ๐‘ก๐‘–๐‘๐‘Ž๐‘™ ๐‘ ๐‘ก๐‘Ÿ๐‘’๐‘ก๐‘โ„Ž

b. ๐‘ฃ๐‘’๐‘Ÿ๐‘ก๐‘–๐‘๐‘Ž๐‘™ ๐‘๐‘œ๐‘š๐‘๐‘Ÿ๐‘’๐‘ ๐‘ ๐‘–๐‘œ๐‘›

c. โ„Ž๐‘œ๐‘Ÿ๐‘–๐‘ง๐‘Ž๐‘›๐‘ก๐‘Ž๐‘™ ๐‘ โ„Ž๐‘–๐‘“๐‘ก ๐‘ก๐‘œ ๐‘กโ„Ž๐‘’ ๐‘Ÿ๐‘–๐‘”โ„Ž๐‘ก

d. โ„Ž๐‘œ๐‘Ÿ๐‘–๐‘ง๐‘œ๐‘›๐‘ก๐‘Ž๐‘™ ๐‘ โ„Ž๐‘–๐‘“๐‘ก ๐‘ก๐‘œ ๐‘กโ„Ž๐‘’ ๐‘™๐‘’๐‘“๐‘ก

e. ๐‘ฃ๐‘’๐‘Ÿ๐‘ก๐‘–๐‘๐‘Ž๐‘™ ๐‘ โ„Ž๐‘–๐‘“๐‘ก ๐‘ข๐‘

f. vertical shift down

Solution :

Since the value of a lies between 0 to 1, vertical compression can be done. So, the answer is option b.

Problem 9 :

If โ„Ž > 0 what would be the transformation of, โ„Ž , in the vertex form of the quadratic equation ๐‘ฆ = ๐‘Ž(๐‘ฅ โˆ’ โ„Ž) 2 + ๐‘˜?

a. ๐‘ฃ๐‘’๐‘Ÿ๐‘ก๐‘–๐‘๐‘Ž๐‘™ ๐‘ ๐‘ก๐‘Ÿ๐‘’๐‘ก๐‘โ„Ž

b. ๐‘ฃ๐‘’๐‘Ÿ๐‘ก๐‘–๐‘๐‘Ž๐‘™ ๐‘๐‘œ๐‘š๐‘๐‘Ÿ๐‘’๐‘ ๐‘ ๐‘–๐‘œ๐‘›

c. โ„Ž๐‘œ๐‘Ÿ๐‘–๐‘ง๐‘Ž๐‘›๐‘ก๐‘Ž๐‘™ ๐‘ โ„Ž๐‘–๐‘“๐‘ก ๐‘ก๐‘œ ๐‘กโ„Ž๐‘’ ๐‘Ÿ๐‘–๐‘”โ„Ž๐‘ก

d. โ„Ž๐‘œ๐‘Ÿ๐‘–๐‘ง๐‘œ๐‘›๐‘ก๐‘Ž๐‘™ ๐‘ โ„Ž๐‘–๐‘“๐‘ก ๐‘ก๐‘œ ๐‘กโ„Ž๐‘’ ๐‘™๐‘’๐‘“๐‘ก

e. ๐‘ฃ๐‘’๐‘Ÿ๐‘ก๐‘–๐‘๐‘Ž๐‘™ ๐‘ โ„Ž๐‘–๐‘“๐‘ก ๐‘ข๐‘

f. vertical shift down

Solution :

Horizontal shift to the right.

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