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

• Feb 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!
• Feb 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 +).
• Feb 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?
• Feb 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.
• Feb 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.
• Feb 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.
• Feb 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 (multiplication/division).

This seems wrong.

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?
• Feb 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 (multiplication/division).

This seems wrong.