difference between multiplying y by a coefficient and x by a coefficient

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February 23rd 2011, 11:12 AM
jonnygill

difference between multiplying y by a coefficient and x by a coefficient

Hello,

So lets say we had the equation $g(x)=3(x+1)^2$

this is basically the parent function, $f(x)=x^2$ , but shifted to the left 1 unit and expanded vertically by a factor of 3. The coefficient 3 is essentially being multiplied by f(x) aka y which could be represented by the expression $3\cdot f(x)$ . Since what is done to one side of an equation must be done to the other side, we must multiply (x^2) by 3 as well, or in this case, $3(x+1)^2$ .

So, i feel like i understand this. What i don't understand is, why when we are multiplying f(x) by 3 we say "...by a factor of three" and when we multiply x by 3 we say "...by a factor of one third." I don't see where the 1/3 comes from algebraically and it isn't explained in the book i am reading, it just expects the reader to accept it as a fact. Well, i accept it, but i still want to know where it comes from.

Thanks!
February 23rd 2011, 11:45 AM
emakarov

What you probably mean is that the graph of f(3x) is the graph of f(x) expanded horizontally by a factor 1/3, which is the same thing as contracted by a factor of 3. If $g(x) = f(3x)$ and $f(x_0)=y_0$ for some particular $x_0$ , then $g(x_0/3)=y_0$ , which shows that the graph is squeezed, rather than expanded, horizontally. This is analogous to the fact that the graph of f(x + 1) is the graph of f(x) shifted to the left (despite using +).
February 23rd 2011, 12:26 PM
jonnygill

Quote:

Originally Posted by emakarov

What you probably mean is that the graph of f(3x) is the graph of f(x) expanded horizontally by a factor 1/3, which is the same thing as contracted by a factor of 3. If $g(x) = f(3x)$ and $f(x_0)=y_0$ for some particular $x_0$ , then $g(x_0/3)=y_0$ , which shows that the graph is squeezed, rather than expanded, horizontally. This is analogous to the fact that the graph of f(x + 1) is the graph of f(x) shifted to the left (despite using +).

i don't know what $x_0$ represents

isn't it like initial or something?
February 23rd 2011, 12:45 PM
emakarov

You can ignore the subscript. I denoted some particular value by $x_0$ to emphasize that it is kept fixed in the discussion that follows in contrast to x, which may range over the whole function domain and potentially assume different values. Thus, I may write f(x) to mean the whole function f, but $f(x_0)$ to mean a specific number -- the value of f at $x_0$ . Basically, I chose chose some x and gave it a specific name that wouldn't change in the following five minutes.
February 23rd 2011, 12:57 PM
jonnygill

Are you absolutely sure that f(3x) is the graph of f(x) EXPANDED horizontally by a factor of 1/3 and not COMPRESSED horizontally by a factor of 1/3?

For example, if i compress a standard looking letter "U" horizontally, we'll say the "U" corresponds to the graph of the parent function y=x^2, then it seems that that U will get skinnier/thinner/narrower/etc. it would be like taking two bulldozers approaching the U from the left and the right and compressing it inwards. In this case, imagining the graph changing, the output for x=3 would go from y=9 to y>9.

I don't understand the notation you used in your explanation, perhaps if i did this would become clear to me.

The analogy you made makes sense to me only superficially. What i mean by this is that I don't really understand why x+1 moves a graph to the left. So, while it makes sense that if adding 1 moves to the left, then multiplying by 3 would... shoot this isn't adding up.

ok, lets say that y=(3x)^2

if x=3 then y=81

lets now look at x=3 in the parent function... y=x^2.

if x=3 in the parent function then y=9.

So y went from 9 to 81 in this example when we multiplied x by 3.

Imagining a graph changing from y=x^2 to y=(3x)^2 we would see that the graph is being COMPRESSED horizontally. Expanding it horizontally would result in lesser y values. But when we multiply x by a constant c that is >1, the y-values increase, thus the graph is compressed horizontally.

this is frustrating for me because i see that emakarov makes a good point with the whole analogous thing.

p.s. if emakarov is entirely correct, the math book i am looking at has confused expanding horizontally with compressing horizontally in every single problem. like, everytime x is multiplied by a coefficient that is greater than 1 the answer in the back of the book says compressed horizontally by a factor of 1/c.

if someone were to clarify this madness i would feel better i think.
February 23rd 2011, 01:34 PM
emakarov

In fact, this is one of the most confusing issues in math for me. I understand it pretty well, but every time I have to decide whether to multiply or divide to compress a graph, whether to add or subtract to move a graph to the left, I have to think long and hard. This is also pretty similar to daylight saving time, where we move the clock's hand forward in order to get up earlier; I can never remember in which direction I should update the clock.

Quote:

Imagining a graph changing from y=x^2 to y=(3x)^2 we would see that the graph is being COMPRESSED horizontally.

Precisely. The value y = 81 corresponds to x = 9 for the function y = x^2, but y = 81 corresponds to x = 3 for y = (3x)^2. This means that the point with the same vertical coordinate is located 3 times closer to the Y axes in the second case, i.e., the graph is compressed horizontally when we move from x^2 to (3x)^2.

Quote:

Are you absolutely sure that f(3x) is the graph of f(x) EXPANDED horizontally by a factor of 1/3 and not COMPRESSED horizontally by a factor of 1/3?

Note that I said "expanded by a factor of 1/3". Since the factor is < 1, it actually means compressed three times.

Quote:

What i mean by this is that I don't really understand why x+1 moves a graph to the left.

Suppose that f(0) = 5 for some function f and let g(x) = f(x + 1). What value of x do you have to give to g in order to get the same output 5? You have to give x = -1, i.e., g(-1) = f(-1 + 1) = f(0) = 5. Thus, in the graph of f(x) we have a point (0, 5). In the graph of g(x) we have a point (-1, 5). Every point of the graph of f(x) is shifted to the left by 1.

Quote:

it would be like taking two bulldozers approaching the U from the left and the right and compressing it inwards.

Nice illustration :)

Quote:

I don't understand the notation you used in your explanation, perhaps if i did this would become clear to me.

$x_0$ is just another name, similar to $x$ , $y$ , $z$ , $a$ , $b$ , $u$ , etc. The subscript does not mean that I apply some operation to $x$ . You can replace it with $a$ if you'd like.
February 23rd 2011, 02:14 PM
jonnygill

Quote:

Note that I said "expanded by a factor of 1/3". Since the factor is < 1, it actually means compressed three times.

So, expanding horizontally by a factor of 1/3 is the same as compressing horizontally by a factor of 3?

If this is true, the math book i am using has it half right which is another way of saying completely wrong. Depending on how you look at it, they are either switching the word expand with compress or 1/3 with 3. I have checked several odd numbered problems (which have answers in the back) and if x if multiplied by a coefficient c where c>1, they will say that the parent graph has been compressed horizontally by a factor of 1/c. When in fact, the graph is either compressed horizontally by a factor of c, or expanded horizontally by a factor of 1/c. The book correctly explains what happens when you multiply f(x) by a constant c. They say that if f(x) is multiplied by a constant c where c>1, then the graph of f(x) is expanded vertically by a factor of c. They also correctly explain that when c is added to x where c>0 the graph will shift to the left c units (and vice-versa).

But there's one thing that "doesn't add up" and it seems to be "logic" based...

if adding c to x where c>0 results in a horizontal shift to the left c units (aka horizontal shift of -c units)

and

multiplying c by x where c>1 results in a horizontal compression by a factor of c (aka horizontal compression of c units)

then observe the above statements in the parentheses.

c becomes -c (addition/subtraction)

c becomes c (multiplication/division).

This seems wrong.

If c becomes -c (addition/subtraction)

then c should becomes 1/c (multiplication/division).

note: i realize this may be difficult to understand, but i'm having a difficult time expressing the logical contradiction that *appears* to be occurring here.

anyone?
February 24th 2011, 12:59 AM
emakarov

Quote:

So, expanding horizontally by a factor of 1/3 is the same as compressing horizontally by a factor of 3?

Yes.

Quote:

I have checked several odd numbered problems (which have answers in the back) and if x if multiplied by a coefficient c where c>1, they will say that the parent graph has been compressed horizontally by a factor of 1/c.

This is strange. Perhaps I would not be too critical if they used some terminology, even nonintuitive, consistently. Mathematics is notorious for redefining common words like "almost everywhere" and "meager" and giving them precise meaning.

I use the phrase "expand horizontally by a factor of c" to mean that each point (x, y) of the function's graph is mapped into (cx, y), i.e., the x coordinate is multiplied by c. Similarly, "compress horizontally by a factor of c" means mapping (x, y) to (x/c, y). In contrast, perhaps your when your textbook says "by a factor of c", it always multiplies the x coordinate by c. In addition, when c > 1, it says "expand" and when c < 1, it says "compress".

The main point is consistency. But if they refer to the same transformation both as "compressing by a factor of 1/2" and "compressing by a factor of 2", or refer to different transformations by the same name, then this is a problem.

Quote:

But there's one thing that "doesn't add up" and it seems to be "logic" based...

if adding c to x where c>0 results in a horizontal shift to the left c units (aka horizontal shift of -c units)

and

multiplying c by x where c>1 results in a horizontal compression by a factor of c (aka horizontal compression of c units)

then observe the above statements in the parentheses.

c becomes -c (addition/subtraction)

c becomes c (multiplication/division).

This seems wrong.

If c becomes -c (addition/subtraction)

then c should becomes 1/c (multiplication/division).

First, a logical contradiction must be much more concrete. If you proved that 0 = 1, that would be a contradiction. As it is, there are some nonintuitive things in math. Here is a quote from an article I read just yesterday.

Quote:

Jerry Bona once said,

The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?

This is a joke. In the setting of ordinary set theory, all three of those principles are mathematically equivalent -- i.e., if we assume any one of those principles, we can use it to prove the other two. However, human intuition does not always follow what is mathematically correct. The Axiom of Choice agrees with the intuition of most mathematicians; the Well Ordering Principle is contrary to the intuition of most mathematicians; and Zorn's Lemma is so complicated that most mathematicians are not able to form any intuitive opinion about it.

Such failing of intuition is not necessarily a contradiction.

Here, in fact, the analogy between addition and multiplication is pretty accurate. When we apply a transformation to the argument of a function, the function's graph undergoes the inverse transformation horizontally. When f(x) is changed into f(x + c), each point (x, y) of the graph is mapped to (x - c, y). When f(x) is changed into f(cx), each point (x, y) of the graph is mapped into (x/c, y). In the first case, + c becomes - c, in the second case, * c becomes / c.