Exploring Families of Functions Notes + Study Guide

General rules for transforming functions:

You can typically write functions as

a(bx – h) + k

Where “a”, “b”, “h”, and “k” are constants.

I. “a” represents the constant responsible for a vertical stretch or compression.

If |a| ≥1 (if “a” is not a fraction that’s less than 1) then the graph will be vertically

stretched.

If |a| ≤ 1 (if “a” is a fraction that’s less than 1 like 2

3 𝑜𝑟

7

8 for example) then the graph

will be vertically compressed.

In both cases, the y coordinates are multiplied by a, but when the graph is vertically

stretched it gets bigger in the y direction. When the graph is vertically compressed it

gets smaller in the y direction. You can tell this from the names “stretched” and

“compressed” too, but here are some sample graphs to show you what changing the

value of “a” can do:

In graph A, the red graph is the parent function y = (x-2) and the blue graph is the

modified function y = 2(x-2). So there is a vertical stretch of 2. You can see this

because for the same x coordinate, the y value of the blue line is twice that of the red

line. Take x = 10 for example. The y coordinate for the red line is 8, but the

GRAPH A GRAPH B

y coordinate for the blue line is 16 which = 2 x 8. So vertically stretching the graph

by 2 means all of the y coordinates get multiplied by 2.

In graph B, the red graph is still the parent function y = (x-2) but the blue graph is

the modified function y = 0.5(x-2). So there is a vertical compression by 0.5 since

|0.5| ≤ 1. Thus each y coordinate is multiplied by 0.5 and gets smaller. Take x = 10

as your sample point again. In the red graph, the y coordinate is 8 but in the blue

graph the y coordinate is 4 which equals 0.5 x 8.

If you get a question like this on a test and aren’t sure whether to shrink or

compress the function, you can always pick a value for x and solve for y like this:

So since the new y value is 2 x the old one, the function is stretched vertically.

If “a” is negative, then the graph is reflected over the x axis. This is because the y

coordinate changes to its opposite sign when it is multiplied by a negative. So if you

had a point (x, y), reflecting it over the x axis makes the new point (x, -y). We can see

this by graphing as well.

GRAPH C

In graph C, the red graph is the same

parent function y = (x-2) but the blue

graph is the modified function

y = -(x-2). “a” in this case is -1. So you

can see that each y coordinate in the red

graph is just the negative value for the

blue graph. Take a sample point again, x

= 10 for example. In the red graph, you

have the point (10, 8) so the y

coordinate = 8. In the blue graph

however, you have the point (10, -8) so

the y coordinate is -8. In other words,

each red y coordinate is simply

multiplied by “–a” (-1 in this case) to get

the blue y coordinate. So you’re still just

multiplying each y value by “a”, but since

“a” is negative here, you reflect each

point over the x axis.

FOR PARENT FUNCTION

y = (x - 2)

y = (10 – 2) y = 8

FOR MODIFIED FUNCTION

y = 2(x - 2)

y = 2(10 – 2)

y = 20 – 4 y = 16

GRAPH D GRAPH E

II. “b” represents the constant responsible for a horizontal stretch or compression. If |b| ≥1 (if “b” is not a fraction that’s less than 1) then the graph will be

horizontally compressed.

If |b| ≤ 1 (if “b” is a fraction that’s less than 1 like 2

3 𝑜𝑟

7

8 for example) then the graph

will be horizontally stretched.

This is more or less the opposite of what happens when |a| is greater than or less than 1. (If |a| ≥ 1, the graph is vertically stretched, but if |b| ≥ 1, the graph is

horizontally compressed.)

In both horizontal stretches and compressions, the new x coordinates are just the

old ones multiplied by the reciprocal of “b.” When the graph is horizontally

stretched it gets bigger in the x direction.

When the graph is horizontally compressed it gets smaller in the x direction.

Again, here are some graphs to show you what changing the value of “b” does”

In graph E, the red graph is the same parent function y = (x - 2) and the blue graph is

the modified function y = (2x – 2). So there is a horizontal compression by 2 in the

blue graph. This means that the x values in the blue graph are ½ the x values in the

red graph for the same y value. Thus the new x coordinates are the old ones

multiplied by the RECIPROCAL of 2, which is ½ . Take the sample point, y = 8. In the

red graph, if you find y = 8 and look at the x value, you can see the point is (10, 8) so

the x coordinate is 10. Do the same for the blue graph. At y = 8, x = 5, which equals

0.5 x 10 or ½ the x value from the red graph. So the graph shrinks horizontally. If

you get a question like this on a test and aren’t sure whether to shrink or compress

the function, you can always pick a value for y and solve for x like this.

FOR MODIFIED FUNCTION

y = (2x - 2)

8 = (2x – 2)

2x = 8 + 2

2x = 10

x = 5

x

FOR PARENT FUNCTION

y = (x - 2)

8 = (x – 2)

x = 8 + 2

x = 10

So you can tell that since the new x value is ½ the old one, the graph shrinks.

In graph B, the red graph is still the parent function y = (x-2) but the blue graph is

the modified function y = (0.5x-2). So there is a horizontal stretch since |0.5| ≤ 1.

Thus each x coordinate is multiplied by 2 (multiplied by the reciprocal of ½) and

gets bigger. Take y = 8 as your sample point again. In the red graph, the x coordinate

is 10 but in the blue graph the x coordinate is 20, which equals 2 x 10.

If “b” is negative, the graph is reflected over the y axis. This is because its x

coordinate changes to a negative when it is multiplied by its opposite sign, and in

order to do this, the points are reflected over the y axis. We can see this by graphing

as well.

In graph C, the red graph is the same

parent function y = (x-2) but the blue

graph is the modified function

y = (-x-2). “b” in this case is -1. So you

can see that each x coordinate in the red

graph is just the negative value for the

blue graph. Take a sample point again, y

= 2 for example. In the red graph, you

have the point (4, 2) so the y coordinate

= 4. In the blue graph however, you have

the point (-4, 2) so the x coordinate is -4.

In other words, each red x coordinate is

simply multiplied by “–b” (-1 in this

case) to get the blue x coordinate. So

you’re multiplying each x value by “b”,

but since “b” is negative here, you reflect each point over the y axis.

GRAPH F GRAPH G

FOR PARENT FUNCTION

y = (x)

First pick a y value on the graph… I

chose y = 0.

0 = x So the graph intercepts the y axis at x = 0

FOR MODIFIED FUNCTION

y = (x - 2)

Pick the same y value you chose for the

other one. In this case, it’s still y = 0

0 = (x – 2)

X = 2 So the graph intercepts the y axis at x = 2

III. “h” is the constant responsible for a horizontal phase shift (for moving the

function left or right).

(x – h) moves the parent function y = x “h” units right.

(x + h) moves the parent function y = x “h” units left.

While this might seem counterintuitive, we can prove this by graphing. Here are

some examples of how adding or subtracting a constant “h” changes a function:

In graph F, the red graph is the parent function y = x and the blue graph is the

modified function y = (x – 2). You can see that the blue graph is the same as the red

one except that is it moved 2 units to the right.

In graph G, the red graph is the parent function y = x and the blue graph is the

modified function y = (x + 2). You can see that the blue graph is the same as the red

one except that is it moved 2 units to the left.

If you get a question like this on a test and aren’t sure if the function moves to the

right or left, you can find a value of y that is in both graphs and try to find the x

values for them like this:

Since x = 2 is two units to the right from x = 0, the function moves 2 to the right

GRAPH H GRAPH I

IV. “k” is the constant responsible for a vertical phase shift (for moving the function

up or down).

x - k moves the parent function y = x “k” units down.

x + k moves the parent function y = x “k” units up.

We can prove this by graphing. Here are some examples of how adding or

subtracting a constant “k” changes a function:

In graph H, the red graph is the parent function y = (2x +1)2 and the blue graph is

the modified function y = (2x + 1)2 – 3. So you can see that the blue function is the

same as the red one but is shifted 3 units down.

In graph I, the red graph is the parent function y = (2x +1)2 and the blue graph is the

modified function y = (2x + 1)2 + 3. The blue function is the same as the red one but

is shifted 3 units up.

The key difference between the constants “h” and “k” is that “h” is applied only to x

whereas “k” is applied to the whole function (“h” is inside of the parentheses

containing x and “k” is outside.)

So the function (2x + 1)2 – 3 has a horizontal phase shift 1 unit to the left and a

vertical phase shift 3 units down.

GRAPH J GRAPH K

GRAPH L

V. Domain:

Domain is basically all of the x values that a function can be defined for (all of the x

values a function can have). Here are some examples of graphs and their domains.

Graph J is the function y = x. If we were to make the window of the graph huge,

(-10,000,000 to 10,000,000), for example, we would see that the graph still continues in both directions. Thus, since the x values can extend from −∞ to ∞, the domain of the

function is (−∞, ∞).

Graph K is the function y = x2. You can plug in any number for x so the domain is still (−∞, ∞).

In graph L, the function is

Y = 𝑥

𝑥2+3𝑥−4

We can factor the denominator into

(x +4)(x – 1) which means that x

cannot equal -4 or 1. Because of this,

there are vertical asymptotes at -4,

and 1 as shown by the dotted green

lines. Thus, there are no defined

values for x at -4 and 1.

The domain is

(−∞, −4) 𝑈 (−4, −1) 𝑈 (1, ∞)

GRAPH J GRAPH K

GRAPH L

VI. Range:

Range is basically all of the y values that a function can be defined for (all of the y

values a function can have). The same examples of graphs and their domains can be

used to find their ranges:

The range of Graph J is (-∞ , ∞) since there are negative values for y and positive

values.

The range of Graph K is (0, ∞) because there are no y values below 0 and there can’t

be any below 0. If you try plugging any negative x into the function y = x 2 you will

get a positive number for y. Thus, there cannot be any negative values and the range

begins at 0.

In graph L, the function is

Y = 𝑥

𝑥2+3𝑥−4

The range is (−∞, ∞)

Because there are y values that stretch in both directions.

VII. CHEAT SHEET:

For transformations of functions that follow the format a(bx – h) +k:

1. If |a| ≥ 1, function is vertically stretched

2. If |a| ≤ 1, function is vertically compressed

3. If |b| | ≥ 1, function is horizontally compressed

4. If |b| ≤ 1, function is horizontally stretched

5. If h is +

(x – (+h)) = (x – h)

function moves to the right

6. If h is –

(x – (-h)) = (x + h)

Function moves to the left

7. If k is + function moves up

8. If k is – function moves down

Domain and Range: 1. For even powered functions, domain = (−∞, ∞)

* Range = (0, ∞) UNLESS THERE IS A VERTICAL PHASE SHIFT

2. For odd powered functions, domain = (−∞, ∞)

Range = (−∞, ∞)

3. For absolute value functions, domain = (−∞, ∞)

* Range = (0, ∞) UNLESS THERE IS A VERTICAL PHASE SHIFT

* IF there is a vertical phase shift, the 0 will be replaced with the minimum y value the

function can have.