2.5 translations of graphs

Transformations of Graphs

http://www.lahc.edu/math/precalculus/math_260a.html





Function Calisthenics (Origin Unknown)

Objective:

* Vertical stretches and compressions of graphs

* Vertical and horizontal translations of graphs


Using image manipulation software, we can drag and drop or stretch images.


Using image manipulation software, we can drag and drop or stretch images. For example, from the original image below,


y = f(x)

Using image manipulation software, we can drag and drop or stretch images. For example, from the original image below, we can elongate it,

stretch


y = f(x)

Using image manipulation software, we can drag and drop or stretch images. For example, from the original image below, we can elongate it, drag and lower it,

stretch

lower


y = f(x)


stretch

lower

vertically reflected

and reflect it vertically to create another pattern.


y = f(x)


stretch

lower

vertically reflected

and reflect it vertically to create another pattern.

If the original image is the graph of the function y = f(x), then these transformations can be tracked easily with the notation of functions.


y = f(x)

Given a function y = f(x) and P = (x, y = f(x)) a generic point on the graph as shown, the output f(x) = y represents the height of the point.

x

P = (x, f(x))

f(x) = ht y= f(x)

x

Vertical Translations

Given a function y = f(x) and P = (x, y = f(x)) a generic point on the graph as shown, the output f(x) = y represents the height of the point. Hence expressions in terms of f(x) may be translated precisely intothe corresponding manipulation of the graph.

x

P = (x, f(x))

f(x) = ht y= f(x)

x


Given a function y = f(x) and P = (x, y = f(x)) a generic point on the graph as shown, the output f(x) = y represents the height of the point. Hence expressions in terms of f(x) may be translated precisely intothe corresponding manipulation of the graph.Vertical Translations x

P = (x, f(x))

f(x) = ht y= f(x)

x



x

P = (x, f(x))

f(x) = ht y= f(x)

Hence expressions in terms of f(x) may be translated precisely intothe corresponding manipulation of the graph.Vertical TranslationsChanging the y–coordinate to f(x) + 3 moves P vertically up 3 units.

(x, f(x)+3)

x



x

P = (x, f(x))

f(x) = ht y= f(x)

Hence expressions in terms of f(x) may be translated precisely intothe corresponding manipulation of the graph.Vertical TranslationsChanging the y–coordinate to f(x) + 3 moves P vertically up 3 units.

(x, f(x)+3) y= f(x) + 3

Hence setting y = f(x) + 3 to all the points on the graph means to move the entire graph 3 units up as shown.

x



x

P = (x, f(x))

f(x) = ht y= f(x)

Hence expressions in terms of f(x) may be translated precisely intothe corresponding manipulation of the graph.Vertical TranslationsChanging the y–coordinate to f(x) + 3 moves P vertically up 3 units.Hence setting y = f(x) + 3 to all the points on the graph means to move the entire graph 3 units up as shown. Likewise changing the y–coordinate to f(x) – 3 corresponds to moving y = f(x) down 3 units.

(x, f(x)+3)

(x, f(x)–3)

y= f(x) – 3

y= f(x) + 3

x


The graph of (x, y = f(x) + c) is the vertical translation of the graph (x, f(x)).

Vertical TranslationsVertical Translations



P = (x, f(x)) y= f(x)

(x, f(x)+c) where c > 0

(x, f(x)+c) where c < 0




P = (x, f(x)) y= f(x)


(x, f(x)+c) where c < 0 Here are the graphs of:

y = f(x) = x2 vs. y = f(x) + 5 = x2 + 5

y = x2

y = x2 + 5

(0, 0)

(0, 5)



P = (x, f(x)) y= f(x)



y = f(x) = x2 vs. y = f(x) + 5 = x2 + 5

y = x2

y = x2 + 5

(0, 0)

(0, 5)

x


The graph of (x, y = f(x) + c) is the vertical translation of the graph (x, f(x)). Assuming c > 0, move the graph (x, f(x)) up to obtain the graph (x, f(x) + c).


P = (x, f(x)) y= f(x)



y = f(x) = x2 vs. y = f(x) + 5 = x2 + 5

y = f(x) = x2 vs. y = f(x) – 5 = x2 – 5

y = x2

y = x2 + 5

y = x2 – 5

y = x2

(0, 0)

(0, 5)

(0, 0)

(0, –5)

xx





move the graph (x, f(x)) down to obtain the graph (x, f(x) – c).

P = (x, f(x)) y= f(x)



y = f(x) = x2 vs. y = f(x) + 5 = x2 + 5

y = f(x) = x2 vs. y = f(x) – 5 = x2 – 5

y = x2

y = x2 + 5

y = x2 – 5

y = x2

(0, 0)

(0, 5)

(0, 0)

(0, –5)

xx


Given a function y = f(x) and P = (x, y = f(x)) a generic point on the graph as shown.

Vertical Stretches and Compressions

P = (x, f(x)) y= f(x)


Given a function y = f(x) and P = (x, y = f(x)) a generic point on the graph as shown. Changing the y–coordinate to 3f(x)would triple the height of the point P.


P = (x, f(x)) y= f(x)




P = (x, f(x)) y= f(x)

Hence setting y = 3f(x) to all the points on the graph means to elongate the entire graph 3 times while the x–intercepts (x, 0)’s remain fixed because 3(0) = 0.

y= 3f(x)




P = (x, f(x)) y= f(x)


Likewise setting y = (1/3)f(x) would compress the entire graph to a third of it’s original size while the x–intercepts or (x, 0)’s remain fixed.

y= 3f(x)




P = (x, f(x)) y= f(x)


Likewise setting y = (1/3)f(x) would compress the entire graph to a third of it’s original size while the x–intercepts or (x, 0)’s remain fixed.

y= 3f(x)

y= f(x)/3


Vertical Stretches and CompressionsAssuming c > 0, the graph of y = cf(x) is the vertical-stretch or compression of y = f(x).


Vertical Stretches and CompressionsAssuming c > 0, the graph of y = cf(x) is the vertical-stretch or compression of y = f(x).

Here are the graphs of: y = f(x) = 4 – x2 vs. y = 3f(x) = 3(4 – x2)

y = 4 – x2

y = 3(4 – x2)

(0, 4)

(0, 12)

(–2, 0) (2, 0) x


Vertical Stretches and CompressionsAssuming c > 0, the graph of y = cf(x) is the vertical-stretch or compression of y = f(x). If c > 1, it is a vertical stretch by a factor of c.


y = 4 – x2

y = 3(4 – x2)

(0, 4)

(0, 12)

(–2, 0) (2, 0)

c = 3

x


Vertical Stretches and CompressionsAssuming c > 0, the graph of y = cf(x) is the vertical-stretch or compression of y = f(x). If c > 1, it is a vertical stretch by a factor of c.


y = f(x) = 4 – x2 vs. y = f(x)/2 = (4 – x2)/2

y = 4 – x2

y = 3(4 – x2)y = 4 – x2

y = (4 – x2)/2

(0, 4)

(0, 12)

(0, 4)

(0, 2)

(–2, 0) (2, 0)

(–2, 0) (2, 0)

c = 3 c = 1/2

xx


Vertical Stretches and CompressionsAssuming c > 0, the graph of y = cf(x) is the vertical-stretch or compression of y = f(x). If c > 1, it is a vertical stretch by a factor of c. If 0 < c < 1, it is a vertical compression by a factor of c.Here are the graphs of: y = f(x) = 4 – x2 vs. y = 3f(x) = 3(4 – x2)

y = f(x) = 4 – x2 vs. y = f(x)/2 = (4 – x2)/2

y = 4 – x2

y = 3(4 – x2)y = 4 – x2

y = (4 – x2)/2

(0, 4)

(0, 12)

(0, 4)

(0, 2)

(–2, 0) (2, 0)

(–2, 0) (2, 0)

c = 3 c = 1/2

xx


Changing the y-coordinate to –f(x) reflects the point P vertically across the x–axis to Q(x, –f(x)) as shown.

P = (x, f(x))

y= f(x) x



P = (x, f(x))

y= f(x)

Q = (x, –f(x))

x



P = (x, f(x))

y= f(x)

Hence setting y = –f(x) to all the points on the graph means to reflect the entire graph vertically across the x–axis.

y= –f(x)

Q = (x, –f(x))

x



P = (x, f(x))

y= f(x)

Hence setting y = –f(x) to all the points on the graph means to reflect the entire graph vertically across the x–axis. Hence settingy = –cf(x) = c*(–f(x)) to all the points means to reflect the entire graph (x, f(x)) vertically, then stretch the reflection by a factor of c.

y= –f(x)

Q = (x, –f(x))

x



P = (x, f(x))

y= f(x)

Hence setting y = –f(x) to all the points on the graph means to reflect the entire graph vertically across the x–axis. Hence settingy = –cf(x) = c*(–f(x)) to all the points means to reflect the entire graph (x, f(x)) vertically, then stretch the reflection by a factor of c.

y= –f(x)

y= –2f(x)

Q = (x, –f(x))

(x, –2f(x))

x



P = (x, f(x))

y= f(x)

Hence setting y = –f(x) to all the points on the graph means to reflect the entire graph vertically across the x–axis. Hence settingy = –cf(x) = c*(–f(x)) to all the points means to reflect the entire graph (x, f(x)) vertically, then stretch the reflection by a factor of c. The order of applying stretching vs. reflecting does not matter, “reflect then stretch” or “stretch then reflect” yields the same result. This is not the case for “stretch” vs. “vertical shift”.

y= –f(x)

y= –2f(x)

Q = (x, –f(x))

(x, –2f(x))

x


Example A. a. Given the graph of y = f(x),graph y = g(x) = –2f(x) + 3

(–3, 1)

(–1, –1)

(1, 1) (2, 1)

x



(–3, 1)

(–1, –1)

(1, 1) (2, 1)

“–2f(x)" corresponds stretching the graph by a factor of 2, then reflecting the entire graph across the x axis. Afterwards move the graph vertically up by 3.

x



(–3, 1)

(–1, –1)

(1, 1) (2, 1)

“–2f(x)" corresponds stretching the graph by a factor of 2, then reflecting the entire graph across the x axis. Afterwards move the graph vertically up by 3.To draw the graph, track the “important points” on the graph.

x



(–3, 1)

(–1, –1)

(1, 1) (2, 1)

“–2f(x)" corresponds stretching the graph by a factor of 2, then reflecting the entire graph across the x axis. Afterwards move the graph vertically up by 3.To draw the graph, track the “important points” on the graph. Plug in their x–values to find the corresponding y–values, then plot their destinations.

x



(–3, 1)

(–1, –1)

(1, 1) (2, 1)


(–3, 1)

x


y = g(x) = –2f(x) + 3y = f(x)


(–3, 1)

(–1, –1)

(1, 1) (2, 1)


(–3, 1) (–3, –2f(–3) + 3 = 1)

x


y = g(x) = –2f(x) + 3y = f(x)


(–3, 1)

(–1, –1)

(1, 1) (2, 1)


(–3, 1) (–3, –2f(–3) + 3 = 1)(–1, –1) (–1, –2f(–1) + 3 = 5)

x


y = g(x) = –2f(x) + 3y = f(x)


(–3, 1)

(–1, –1)

(1, 1) (2, 1)


(–3, 1) (–3, –2f(–3) + 3 = 1)(–1, –1) (–1, –2f(–1) + 3 = 5)(1, 1) (1, –2f(1) + 3 = 1)(2, 1) (2, –2f(2) + 3 = 1)

x


y = g(x) = –2f(x) + 3y = f(x)


(–3, 1)

(–1, –1)

(1, 1) (2, 1)


(–3, 1) (–3, –2f(–3) + 3 = 1)(–1, –1) (–1, –2f(–1) + 3 = 5)(1, 1) (1, –2f(1) + 3 = 1)(2, 1) (2, –2f(2) + 3 = 1)

(–3, 1)

(–1, 5)

(1, 1)(2, 1)

x

x


y = g(x) = –2f(x) + 3y = f(x) Graph of y = g(x)

Horizontal Translationsy= f(x)

x

Let y = f(x) be as shown.

Horizontal Translationsy= f(x)

x

Let y = f(x) be as shown.Let’s define g(x) = f(x + 1), and the goal is to graph g(x).

Horizontal Translations

x

y= f(x)

x

Let y = f(x) be as shown.Let’s define g(x) = f(x + 1), and the goal is to graph g(x).For the point x, the output isy = g(x) = f(x + 1),


x

(x+1, f(x+1))

ht =f(x+1)

y= f(x)

x

Let y = f(x) be as shown.Let’s define g(x) = f(x + 1), and the goal is to graph g(x).For the point x, the output isy = g(x) = f(x + 1),

x+1


x

(x+1, f(x+1))

ht =f(x+1)

y= f(x)

x

Let y = f(x) be as shown.Let’s define g(x) = f(x + 1), and the goal is to graph g(x).For the point x, the output isy = g(x) = f(x + 1), so the point (x, g(x) = f(x +1)) is on the graph of g(x). x+1


x

(x+1, f(x+1))

ht =f(x+1)

y= f(x)

x

Let y = f(x) be as shown.Let’s define g(x) = f(x + 1), and the goal is to graph g(x).For the point x, the output isy = g(x) = f(x + 1), so the point (x, g(x) = f(x +1)) is on the graph of g(x). x+1

(x, f(x +1))


x

(x+1, f(x+1))

ht =f(x+1)

y= f(x)

x

Let y = f(x) be as shown.Let’s define g(x) = f(x + 1), and the goal is to graph g(x).For the point x, the output isy = g(x) = f(x + 1), so the point (x, g(x) = f(x +1)) is on the graph of g(x). ux+1

(x, f(x +1))

Likewise if the input is u,


x

(x+1, f(x+1))

ht =f(x+1)

y= f(x)

x

Let y = f(x) be as shown.Let’s define g(x) = f(x + 1), and the goal is to graph g(x).For the point x, the output isy = g(x) = f(x + 1), so the point (x, g(x) = f(x +1)) is on the graph of g(x). u

(u+1, f(u+1))

ht = f(u+1)

u+1x+1

(x, f(x +1))

Likewise if the input is u, then y = g(u) = f(u + 1)


x

(x+1, f(x+1))

ht =f(x+1)

y= f(x)

x


(u+1, f(u+1))

ht = f(u+1)

u+1x+1

(x, f(x +1))

(u, f(u +1))

Likewise if the input is u, then y = g(u) = f(u + 1) so the point (u, f(u +1)) is on the graph of g(x).

Hence to obtain the graph of y = f(x + 1),shift the entire graph of y = f(x) left by 1 unit.


x

(x+1, f(x+1))

ht =f(x+1)

y= f(x)

x


(u+1, f(u+1))

ht = f(u+1)

u+1x+1

y = g(x) or y = f(x + 1)

(x, f(x +1))

(u, f(u +1))


Shifting left by 1

Hence to obtain the graph of y = f(x + 1),shift the entire graph of y = f(x) left by 1 unit.


x

(x+1, f(x+1))

ht =f(x+1)

y= f(x)

x


(u+1, f(u+1))

ht = f(u+1)

u+1x+1

y = g(x) or y = f(x + 1)

(x, f(x +1))

(u, f(u +1))


Shifting left by 1

Similarly demonstrations show that to obtain the graph of y = f(x – 1), shift the graph of y = f(x) right by 1 unit.

Horizontal Translations The graphs of y = f(x + c) are the horizontal shifts (or translations) of y = f(x) by c units,


Horizontal Translations The graphs of y = f(x + c) are the horizontal shifts (or translations) of y = f(x) by c units,


Example B. Graph the following functions by shiftingthe graph of y = f(x) = x2. Label their vertices.a. g(x) = (x + 2)2 = f(x + 2)

x

y=x 2

b. h(x) = (x – 2)2 = f(x – 2)

Horizontal Shifts

Horizontal Translations The graphs of y = f(x + c) are the horizontal shifts (or translations) of y = f(x) by c units, assuming c > 0:


moves y = f(x) to the left for y = f(x + c).

Example B. Graph the following functions by shiftingthe graph of y = f(x) = x2. Label their vertices.a. g(x) = (x + 2)2 = f(x + 2)

b. h(x) = (x – 2)2 = f(x – 2)x

y=x 2

Horizontal Shifts




b. h(x) = (x – 2)2 = f(x – 2)

Example B. Graph the following functions by shiftingthe graph of y = f(x) = x2. Label their vertices.a. g(x) = (x + 2)2 = f(x + 2)Shift of the graph of y = x2

left 2 units.

x

y=x 2

Horizontal Shifts


y=(x + 2)2




left 2 units.

x

y=x 2

b. h(x) = (x – 2)2 = f(x – 2)

Horizontal Shifts


y=(x + 2)2




left 2 units. The vertex of g(x) = (x + 2)2 is (–2, 0).

x

y=x 2

b. h(x) = (x – 2)2 = f(x – 2)

Horizontal Shifts

(–2, 0)


y=(x + 2)2


moves y = f(x) to the left for y = f(x + c).moves y = f(x) to the right for y = f(x – c). Example B. Graph the following functions by shifting

the graph of y = f(x) = x2. Label their vertices.a. g(x) = (x + 2)2 = f(x + 2)Shift of the graph of y = x2


x

y=x 2

b. h(x) = (x – 2)2 = f(x – 2)

Horizontal Shifts

(–2, 0)


y=(x + 2)2





x

y=x 2

b. h(x) = (x – 2)2 = f(x – 2)Shift the graph of y = x2 right 2 units. The vertex of h(x) is (2, 0). Horizontal Shifts

(–2, 0)


y=(x + 2)2 y=(x – 2)2





x

y=x 2

b. h(x) = (x – 2)2 = f(x – 2)Shift the graph of y = x2 right 2 units. The vertex of h(x) is (2, 0). Horizontal Shifts

(–2, 0) (2, 0)

Horizontal Stretches and Compressions

x

Let y = f(x) with its graph shown here and letg(x) = f(2x).

0

y= f(x)



x

Let y = f(x) with its graph shown here and letg(x) = f(2x).For the point x, the output is g(x) = f(2x)

x0

y= f(x)


x


x

Let y = f(x) with its graph shown here and letg(x) = f(2x).For the point x, the output is g(x) = f(2x)so the point (x, y = f(2x)) is on the graph of g(x).

x0

y= f(x)


x


x

Let y = f(x) with its graph shown here and letg(x) = f(2x).For the point x, the output is g(x) = f(2x)so the point (x, y = f(2x)) is on the graph of g(x).

0

y= f(x)


2x

ht =f(2x)

(2x, f(2x))

x


2x

ht =f(2x) x

Let y = f(x) with its graph shown here and letg(x) = f(2x).For the point x, the output is g(x) = f(2x) so the point (x, y = f(2x)) is on the graph of g(x).

x0

y= f(x)

(2x, f(2x))


(x, g(x)=f(2x))


2x

ht =f(2x) x


x0

y= f(x)

(2x, f(2x))


x

Likewise for the point u, the output is g(u) = f(2u) so the point (u, y = f(2u)) is on the graph of g(x).

(x, g(x)=f(2x))

u


2x

ht =f(2x) x


x0

y= f(x)

(2x, f(2x))


x

2u

(2u, f(2u))

ht = f(2u)


(x, g(x)=f(2x))

u


2x

ht =f(2x) x


x0

y= f(x)

(2x, f(2x))


x

2u

(2u, f(2u))

ht = f(2u)


(x, g(x)=f(2x))

u

(u,g(u))=f(2u)


2x

(u,g(u))=f(2u)

ht =f(2x)

y=g(x)=f(2x)

x

2u

(2u, f(2u))

ht = f(2u)

u

Let y = f(x) with its graph shown here and letg(x) = f(2x).For the point x, the output is g(x) = f(2x) so the point (x, y = f(2x)) is on the graph of g(x).Likewise for the point u, the output is g(u) = f(2u) so the point (u, y = f(2u)) is on the graph of g(x).

x0

y= f(x)

Horizontal stretch by a factor of 2

Hence we see that the graph of y = f(2x) is the horizontal compression of the graph y = f(x) by a factor of ½ .

(x, g(x)=f(2x)) (2x, f(2x))



2x

(u,g(u))=f(2u)

ht =f(2x)

y=g(x)=f(2x)

x

2u

(2u, f(2u))

ht = f(2u)

u

Let y = f(x) with its graph shown here and letg(x) = f(2x).For the point x, the output is g(x) = f(2x) so the point (x, y = f(2x)) is on the graph of g(x).Likewise for the point u, the output is g(u) = f(2u) so the point (u, y = f(2u)) is on the graph of g(x).

x0

y= f(x)

Horizontal stretch by a factor of 2

Hence we see that the graph of y = f(2x) is the horizontal compression of the graph y = f(x) by a factor of ½ . Similarly, the graph of y = f(½ * x) is the horizontal stretch the graph of y = f(x) by a factor of 2. (Convince yourself of this fact.)

(x, g(x)=f(2x)) (2x, f(2x))


Horizontal ReflectionsHorizontal Stretches and Compressions

Let y = f(x) with its graph shown here and letg(x) = f(–x).

x

y= f(x)y

0

Horizontal reflection


Let y = f(x) with its graph shown here and letg(x) = f(–x).For the point x, the output is g(x) = f(–x)

x

y= f(x)

x

y

0



Let y = f(x) with its graph shown here and letg(x) = f(–x).For the point x, the output is g(x) = f(–x)so the point (x, y = f(–x)) is on the graph of g(x).

x

y= f(x)

x

y

–x0




x

y= f(x)

x

y

–x

(x, g(x)=f(–x))

0


Horizontal Reflections

x


0

y= f(x)


x–x

(x, g(x)=f(–x))

u

Likewise for the point u, the output is g(u) = f(–u) so the point (u, y = f(–u)) is on the graph of g(x).

y



x


0

y= f(x)


x–x

(x, g(x)=f(–x))

–uu


y



x


0

y= f(x)


x–x

(x, g(x)=f(–x))

–uu


(u, g(u)=f(–u))

y



x


0

y= f(x)


x–x

(x, g(x)=f(–x))

–uu


(u, g(u)=f(–u))

y


Hence we reflect the graph of y = f(x) horizontally across the y–axis to obtain the graph of y = f(–x).


x


0

y= f(x)


x–x

(x, g(x)=f(–x))

–uu

(u, g(x)=f(–x))


(u, g(u)=f(–u))

y


Hence we reflect the graph of y = f(x) horizontally across the y–axis to obtain the graph of y = f(–x).


x


0

y= f(x)


x–x

(x, g(x)=f(–x))

–uu

(u, g(x)=f(–x))


(u, g(u)=f(–u))

y


Hence we reflect the graph of y = f(x) horizontally across the y–axis to obtain the graph of y = f(–x).To graph y = f(–2x), we compress y = f(x) by a factor of ½ to obtain the graph of y = f(2x), then reflect the result to obtain the graph of y = f(–2x).

Summary of vertical transformations of graphs (c > 0).


Vertical Shifts Vertical Stretches and Compressions

xy= f(x)

y= f(x)

x




xy= f(x)

y= f(x)

y = f(x) + c moves f up by cy = f(x) – c move f down by c

x




y= f(x) + 1

x

y= f(x) + 2

y= f(x)

y= f(x)


x




y= f(x)–1

y= f(x) + 1

x

y= f(x) + 2

y= f(x)–2

y= f(x)–3

y= f(x)

y= f(x)


x




y= f(x)–1

y= f(x) + 1

x

y= f(x) + 2

y= f(x)–2

y= f(x)–3

y= f(x)

y= f(x)


x

c > 1, y = cf(x) stretches f vertically




y= f(x)–1

y= f(x) + 1

x

y= f(x) + 2

y= f(x)–2

y= f(x)–3

y= f(x)

y= 2f(x) y= f(x)

y = f(x) + c moves f up by cy = f(x) – c move f down by c c > 1, y = cf(x) stretches f vertically

x




y= f(x)–1

y= f(x) + 1

x

y= f(x) + 2

y= f(x)–2

y= f(x)–3

y= f(x)

y= 3f(x)

y= 2f(x) y= f(x)


x




y= f(x)–1

y= f(x) + 1

x

y= f(x) + 2

y= f(x)–2

y= f(x)–3

y= f(x)

y= 3f(x)

y= f(x)/3

y= 2f(x) y= f(x)


0 < c < 1, y = cf(x) compresses f

x




y= f(x)–1

y= f(x) + 1

x

y= f(x) + 2

y= f(x)–2

y= f(x)–3

y= f(x)

y= 3f(x)

y= f(x)/3

y= 2f(x)

y= –f(x)

y= f(x)


y = –f(x) reflects f vertically


x




y= f(x)–1

y= f(x) + 1

x

y= f(x) + 2

y= f(x)–2

y= f(x)–3

y= f(x)

y= 3f(x)

y= f(x)/3

y= 2f(x)

y= –f(x) y= –2f(x)

y= –3f(x)

y= f(x)


y = –f(x) reflects f vertically


x

x

–1


Horizontal Shiftsy=f(x)

y=f(x)

–2–3

y


Summary of horizontal transformations of graph (c > 0).

x

x

–1



y=f(x)

–2–3

y


y = f(x + c) moves f left by cy = f(x – c) moves f right by c


x



y=f(x+2) y=f(x+1)




x

x

–1

y=f(x)

–2–3

y



y=f(x+2) y=f(x+1)

y=f(x–2) y=f(x–1)




x

x

–1

y=f(x)

–2–3

y



y=f(x+2) y=f(x+1)

y=f(x–2) y=f(x–1)




x

x

–1

y=f(x)

–2–3

y

c > 1, y = f(cx) compresses f horizontally0 < c < 1, y = f(cx) stretches f horizontally.

–1



y=f(x+2) y=f(x+1)

y=f(x–2) y=f(x–1)

y=f(x) y=f(2x)

–2–3

y


(0,f(0))




x

x

–1



y=f(x+2) y=f(x+1)

y=f(x–2) y=f(x–1)

y=f(x) y=f(2x)

–2–3

y=f(3x) y


(0,f(0))




x

x

–1



y=f(x+2) y=f(x+1)

y=f(x–2) y=f(x–1)

y=f(x)y=f(x/2)

y=f(x/3)

y=f(2x)

–2–3

y=f(3x) y


(0,f(0))




x

x

–1



y=f(x+2) y=f(x+1)

y=f(x–2) y=f(x–1)

y=f(x)y=f(x/2)

y=f(x/3)

y=f(2x)

–2–3

y=f(3x) y


y = f(–x) reflect f horizontally

y=f(–x/3)

(0,f(0))




x

Horizontal TranslationsLet y = f(x) be a function with the interval [0, 1] as its domain as shown.

y

1x

2 3½

y = f(x)

Horizontal TranslationsLet y = f(x) be a function with the interval [0, 1] as its domain as shown.The graph of y = g(x) = f(½ * x) is the horizontal stretch, by a factor of 2, of the graph of y = f(x).

y

1x

2 3½

y = f(x) y=g(x)=f(½ * x)

Horizontal TranslationsLet y = f(x) be a function with the interval [0, 1] as its domain as shown.The graph of y = g(x) = f(½ * x) is the horizontal stretch, by a factor of 2, of the graph of y = f(x). The domain of y = g(x) = f(½ * x) correspondingly is the stretch of the domain of y = f(x), from [0, 1] to [0, 2].

y

1x

2 3½

y = f(x) y=g(x)=f(½ * x)


y

y = f(x)

1x

2 3

The graph of y = g(x) = f(½ * x) is the horizontal stretch, by a factor of 2, of the graph of y = f(x). The domain of y = g(x) = f(½ * x) correspondingly is the stretch of the domain of y = f(x), from [0, 1] to [0, 2].

y=g(x)=f(½ * x)

½

y = f(x/3)

Horizontal TranslationsLet y = f(x) be a function with the interval [0, 1] as its domain as shown.The graph of y = g(x) = f(½ * x) is the horizontal stretch, by a factor of 2, of the graph of y = f(x). The domain of y = g(x) = f(½ * x) correspondingly is the stretch of the domain of y = f(x), from [0, 1] to [0, 2].Similarly the domain of y = h(x) = f(2x) is the compression from [0, 1] to [0, ½ ].

y

1x

2 3½

y = f(x) y=g(x)=f(½ * x)y = f(2x) y = f(x/3)


y

y = f(x)

1x

2 3

The graph of y = g(x) = f(½ * x) is the horizontal stretch, by a factor of 2, of the graph of y = f(x). The domain of y = g(x) = f(½ * x) correspondingly is the stretch of the domain of y = f(x), from [0, 1] to [0, 2].Similarly the domain of y = h(x) = f(2x) is the compression from [0, 1] to [0, ½ ].

y = f(x/3)y=g(x)=f(½ * x)y = f(2x)y = f(3x)

½


y

y = f(x)

1x

2 3

The graph of y = g(x) = f(½ * x) is the horizontal stretch, by a factor of 2, of the graph of y = f(x). The domain of y = g(x) = f(½ * x) correspondingly is the stretch of the domain of y = f(x), from [0, 1] to [0, 2].Similarly the domain of y = h(x) = f(2x) is the compression from [0, 1] to [0, ½ ].

y=g(x)=f(½ * x)The Domain of y = f(cx), c > 0If the domain of y = f(x) is [0, a], then the domain of y = f(cx) is [0, a/c].

y = f(2x)

½

y = f(x/3)y = f(3x)

Horizontal TranslationsExample C. a. Given the graph of the function y = f(x) with the domain [–2, 2], graph y = (x – 3)2 – 1 by applying the transformation rules. Give the new domain and label the end points of the graph.

f(x)=x2

x

(2,4) (–2,4)

2–2 (0,0)


f(x)=x2

x

(2,4) (–2,4)

2–2

i. Shift right 3 units the graph of f(x) = x2 to obtain the graph of y = (x – 3)2.

(0,0)

2–2

Shift right 3 units


f(x)=x2

x

(2,4) (–2,4)

2–2


(0,0)

ii. Lower the graph of y = (x – 3)2 by 1 unit forthe graph of y = (x – 3)2 – 1.

2–2

(5,3) (1,3)

(3,–1)

Shift right 3 units

Lower by

1 unit


f(x)=x2

x

(2,4) (–2,4)

2–2


(0,0)

ii. Lower the graph of y = (x – 3)2 by 1 unit forthe graph of y = (x – 3)2 – 1.

2–2

(5,3) (1,3)

(3,–1)

Shift right 3 units

Lower by

1 unit

The new domain is [–2 + 3, 2 + 3] = [1, 5]. The new vertex is (3, –1) and end points (1, 3) and (5, 3).


x

(4,2)

(0,0)

y=g(x)=√x

4

b. Given the graphs of the function y = g(x) = √x and y = G(x) a transformation of y = g(x) as shown,express G(x) using g(x).

cy=G(x)

(6,2)

62


The graph of y = G(x) is obtained by horizontally compressing the graph of

x

(4,2)

(0,0)

y=g(x)=√x

4


cy=G(x)

(6,2)

62

y = g(x) by a factor of ½, which gives the graph ofh(x) = g(2x) = √2x as shown,

x

(4,2)

(0,0)

y=g(x)=√x

4 62

(2,2)

cy=h(x)=√2x horizontal

compression



x

(4,2)

(0,0)

y=g(x)=√x

4


cy=G(x)

(6,2)

62


x

(4,2)

(0,0)

y=g(x)=√x

4 62

(2,2)

c

then moving h(x) to the right by 4 units.

y=h(x)=√2x

x

(0,0) 4

cy=G(x)

(6,2)

62

(2,2)

cy=h(x)=√2x

horizontal compression

horizontal shift



x

(4,2)

(0,0)

y=g(x)=√x

4


cy=G(x)

(6,2)

62


x

(4,2)

(0,0)

y=g(x)=√x

4 62

(2,2)

c

then moving h(x) to the right by 4 units. HenceG(x) = h(x – 4) = √2(x – 4) or thatG(x) = √2x – 8.

y=h(x)=√2x

x

(0,0) 4

cy=G(x)

(6,2)

62

(2,2)

cy=h(x)=√2x

horizontal compression

horizontal shift

Absolute-Value Flip

Absolute-Value Flip

y = f(x) = x

x -2 -1 0 1

y -2 -1 0 1

Absolute-Value Flip

y = f(x) = x y = |f(x)| = |x|

x -2 -1 0 1

y -2 -1 0 1

x -2 -1 0 1

y 2 1 0 1

Absolute-Value Flip

y = f(x) = x

The graph of y = |f(x)| is obtained by reflecting the portion of the graph below the x-axis to above the x-axis.

y = |f(x)| = |x|

x -2 -1 0 1

y -2 -1 0 1

x -2 -1 0 1

y 2 1 0 1

Absolute-Value FlipAnother example,

y = x2 – 1

(0,–1)


y = x2 – 1 y = |x2 – 1|

(0,–1)


y = x2 – 1 y = |x2 – 1|

(0,–1)

(1,0)


y = x2 – 1 y = |x2 – 1|

y = |x2 – 1| – 1

(0,–1)

(1,0)


y = x2 – 1 y = |x2 – 1|

y = |x2 – 1| – 1

(0,–1)

(0,0)

(1,0)


y = x2 – 1 y = |x2 – 1|

y = |x2 – 1| – 1 y = 2(|x2 – 1| – 1)

(0,–1)

(0,0)

(1,0)


y = x2 – 1 y = |x2 – 1|

y = |x2 – 1| – 1 y = 2(|x2 – 1| – 1)

(0,–1)

(0,0) (0,0)

(1,0)

Horizontal FlipThe graph of y = f(–x) is the horizontal reflection of the graph of y = f(x) across the y axis.


y = f(x) = x3 – x2


y = f(x) = x3 – x2

y = f(-x) = (-x)3 – (-x)2


y = f(x) = x3 – x2

y = f(-x) = (-x)3 – (-x)2

y = f(-x) = – x3 – x2


y = f(x) = x3 – x2

y = f(-x) = (-x)3 – (-x)2

y = f(-x) = – x3 – x2

A function is said to be even if f(x) = f(– x). Graphs of even functions are symmetric to the y-axis.


y = f(x) = x3 – x2

y = f(-x) = (-x)3 – (-x)2

y = f(-x) = – x3 – x2

A function is said to be even if f(x) = f(– x). Graphs of even functions are symmetric to the y-axis.

Graph of an even function

x

(x, f(x))

–x

(–x, f(–x))

Horizontal Flip

Polynomial-functions whose terms are all even powers are even. The graph on the right is the even polynomial–function y = x4 – 4x2. y = x4 – 4x2

Horizontal Flip

A function is said to be odd iff f(–x) = – f(x).


Horizontal Flip

y = x4 – 4x2

A function is said to be odd iff f(–x) = – f(x). Graphs of odd functions are symmetric to the origin,

Polynomial-functions whose terms are all even powers are even. The graph on the right is the even polynomial–function y = x4 – 4x2.

Horizontal Flip

A function is said to be odd iff f(–x) = – f(x). Graphs of odd functions are symmetric to the origin, that is, they're the same as reflecting across the y-axis followed by reflecting across the x-axis.


Horizontal Flip

Graph of an odd function


x–x0

(x, f(x))

(–x, –f(x))


Horizontal Flip



x–x0

(x, f(x))

(–x, –f(x))

u

(u, f(u))

(–u, –f(u))

–u


Horizontal Flip




x–x0

(x, f(x))

(–x, –f(x))

u

(u, f(u))

(–u, –f(u))

–u

Horizontal Flip

Polynomial-functions whose terms are all odd powers are odd.

Horizontal Flip

y = x3 – 4x

Polynomial-functions whose terms are all odd powers are odd. The graph on the right is the odd polynomial–function y = x3 – 4x.

Horizontal Flip

Polynomial-functions whose terms are all odd powers are odd. The graph on the right is the odd polynomial–function y = x3 – 4x. y = x3 – 4x

Theorem (even and odd):

Horizontal Flip

y = x3 – 4x

Theorem (even and odd):I. The sum of even functions is even. The sum of odd functions is odd.


Horizontal Flip

y = x3 – 4x

Theorem (even and odd):I. The sum of even functions is even. The sum of odd functions is odd.II. The product of even functions is even.


Horizontal Flip

y = x3 – 4x

Theorem (even and odd):I. The sum of even functions is even. The sum of odd functions is odd.II. The product of even functions is even. The product of odd functions is even.


Horizontal Flip

y = x3 – 4x

Theorem (even and odd):I. The sum of even functions is even. The sum of odd functions is odd.II. The product of even functions is even. The product of odd functions is even. The product of an even function with an odd function is odd.


Horizontal Flip

y = x3 – 4x

Theorem (even and odd):I. The sum of even functions is even. The sum of odd functions is odd.II. The product of even functions is even. The product of odd functions is even. The product of an even function with an odd function is odd. (The same hold for quotient.)


Horizontal Flip

y = x3 – 4x


is odd, x x4 + 1


Horizontal Flip

y = x3 – 4x


is odd, is even, x x4 + 1

x2 x4 + 1


Horizontal Flip

y = x3 – 4x


is odd, is even, x + 1 is neither.x x4 + 1

x2 x4 + 1


HWUse the graph of y=x2, sketch thegraphs of:

1. y = 3x2

2. y = -2x2

3. y = -0.5x2

4. y=x2 – 1 5. y=2x2 – 16. y= -x2 – 2 7. y=(x+1)2

8. y=(x–3)2

2.5 translations of graphs

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Transcript of 2.5 translations of graphs