Graphing Question (1 Viewer)

axwe7 · Jan 26, 2016

How would you graph y= f(x)??

I mean, for every value of y, wouldn't f(x) be the same? So wouldn't it be a point?

Shuuya · Jan 26, 2016

y is a function of x

So as the x value changes, the y value also changes. The y value is dependent on the changes in the x value.

f(x)=x^2

y=x^2

so when x=2, y=4, when x=10, y=100 and so on

Drsoccerball · Jan 26, 2016

axwe7 said:
How would you graph y= f(x)??

I mean, for every value of y, wouldn't f(x) be the same? So wouldn't it be a point?

What's the question?

Green Yoda · Jan 26, 2016

f(x) stays the same unless its transitioned.
E.g
f(x)+k is shifted up k units
f(x)-k is shifted down k units
f(x-a) is shifted right a units
f(x+a) is shifted left a units

f(x) also changed when it is transformed
E.g
-f(x) is the reflection of f(x) on the x-axis
f(-x) is the reflection of f(x) on the y-axis

leehuan · Jan 26, 2016

Rathin said:
f(x) stays the same unless its transitioned.
E.g
f(x)+k is shifted up k units
f(x)-k is shifted down k units
f(x-a) is shifted right a units
f(x+a) is shifted left a units

f(x) also changed when it is transformed
E.g
-f(x) is the reflection of f(x) on the x-axis
f(-x) is the reflection of f(x) on the y-axis

Here's one to test you

What's f(a-x)?

Shuuya · Jan 26, 2016

leehuan said:
Here's one to test you

What's f(a-x)?

Is that f(x) reflected about y-axis and shifted right a?

axwe7 · Jan 26, 2016

Drsoccerball said:
What's the question?

Search for the question mark ... lolz jk.

How would you graph y=f(x)?

leehuan · Jan 26, 2016

Shuuya said:
Is that f(x) reflected about y-axis and shifted right a?

I consider it as f(-(x-a))

I.e. shift right first THEN reflect

leehuan · Jan 26, 2016

axwe7 said:
Search for the question mark ... lolz jk.

How would you graph y=f(x)?

Do you realise how ambiguous your question is? f(x) could be anything!! x^2? x^5+4x^3? e^x?

Shuuya · Jan 26, 2016

axwe7 said:
Search for the question mark ... lolz jk.

How would you graph y=f(x)?

You need to know what f(x) is to be able to graph it (i.e what is the function)

Shuuya · Jan 26, 2016

leehuan said:
I consider it as f(-(x-a))

I.e. shift right first THEN reflect

Would thinking of it as f((-x)+a) get the same answer?

KingOfActing · Jan 26, 2016

Paradoxica · Jan 26, 2016

KingOfActing said:
View attachment 32830

This graph represents every power function for a sufficiently screwed up scale for y.

Green Yoda · Jan 26, 2016

leehuan said:
Here's one to test you

What's f(a-x)?

It is reflected on the y-axis and moves left a units.

KingOfActing · Jan 26, 2016

Paradoxica said:
This graph represents every power function for a sufficiently screwed up scale for y.

It's an abuse of notation for $y(x) \equiv f(x)$

InteGrand · Jan 26, 2016

Shuuya said:
Would thinking of it as f((-x)+a) get the same answer?

Rathin said:
It is reflected on the y-axis and moves left a units.

$\noindent No. In actuality, the transformation $f(a-x)$ consists of flipping the graph of $y=f(x)$ about the vertical line $x=\frac{a}{2}$. To see why this is true, set $g(x) \equiv f(a-x)$, then we just need to show that $g\left(\frac{a}{2}-\xi\right)=f\left(\frac{a}{2}+\xi\right)$ for any $\xi\in \mathbb{R}$ (this is equivalent to showing that it is indeed a flip about the vertical line in question). And this is very easy to prove.$

$\noindent Alternatively, to think about it Rathin's way, we actually reflect it in the $y$-axis and then shift it \textit{right} by $a$. This is because when we reflect $y=f(x)$'s graph in the $y$-axis, we have the graph of $y=f(-x)$. Now we shift this right by $a$ (so replace the previous $x$ by $x-a$), giving the graph of $y=f(-(x-a))=f(a-x)$, as desired. But personally I find it more intuitive to think about the $f(a-x)$ transform as reflecting the graph of $y=f(x)$ in the vertical line $x=\frac{a}{2}$. This probably makes it easier to sketch too.$

$\noindent Edit: I think Shuuya was thinking of flipping about the $y$-axis first, and then shifting left by $a$. But actually it is the other order (shift left by $a$ first, then flip about $y$ axis), which is correct and equivalent to flipping about $x = \frac{a}{2}$.$

$\noindent In other words, first shifting the graph of $y=f(x)$ left by $a$ gives us the graph of $y = f(x+a)$. Then flipping this about the $y$-axis gives us the graph of $y= f(-x + a)= f(a-x)$ (i.e. just replacing $x$ with $-x$ in the previous equation, as usual for flipping about $y$-axis).$

InteGrand · Jan 26, 2016

$\noindent Here are some common transformations of the graph of $y=f(x)$:$

$$\bullet$ $y=f(x+k)$, $k>0$: shift the graph of $y=f(x)$ \textsl{left} by $k$ units$

$$\bullet$ $y=f(x-k)$, $k>0$: shift the graph of $y=f(x)$ \textsl{right} by $k$ units$

$$\bullet$ $y=f(kx)$, $k>1$: contract the graph of $y=f(x)$ horizontally by a factor of $k$ (if $k<-1$, do the contraction by factor $|k|$ first and then flip about $y$-axis)$

$$\bullet$ $y=f\left(\frac{x}{k}\right)$, $k>1$: \textsl{dilate} the graph of $y=f(x)$ horizontally by a factor of $k$ (if $k<-1$, do the dilation by factor $|k|$ first and then flip about $y$-axis)$

$$\bullet$ $y=f(x)+k$, $k>0$: shift the graph of $y=f(x)$ up by $k$ units$

$$\bullet$ $y=f(x)-k$, $k>0$: shift the graph of $y=f(x)$ \textsl{down} by $k$ units$

$\bullet$ $y=kf(x)$, $k>0$: stretch the graph of $y=f(x)$ vertically a factor of $k$ $\Big{(}$if $0<k<1$, this means you are \textsl{contracting} it vertically by a factor of $\frac{1}{k}$$\Big{)}$

$$\bullet$ $y=f(-x)$: flip the graph of $y=f(x)$ about the $y$-axis$

$$\bullet$ $y=-f(x)$: flip the graph of $y=f(x)$ about the $x$-axis$

$$\bullet$ $y=-f(-x)$: flip the graph of $y=f(x)$ about the $x$-axis and then $y$-axis (or the other order, doesn't matter; this flips the graph about the origin)$

$\bullet$ $y=f(a-x)$: flip the graph of $y=f(x)$ about the vertical line $x=\frac{a}{2}$

$\bullet$ $y= a-f(x)$: flip the graph of $y=f(x)$ about the horizontal line $y=\frac{a}{2}$

$$\bullet$ $x=f(y)$: flip the graph of $y=f(x)$ about the line $y=x$.$

You can play around with these on the http://www.graphsketch.com/ website to get a feel for these. E.g. here's one where we see the effect of $y=f(kx)$ for differing values of k:

http://www.graphsketch.com/?eqn1_co..._lines=1&line_width=4&image_w=850&image_h=525

$\noindent These transformations can also be combined in various ways (order generally being important). For example, to get the graph of $y=f(2x+1)$ from $f(x)$, you can first shift the graph left by 1 to get $y=f(x+1)$, and then contract this horizontally by a factor of 2 to get the graph of $y=f(2x+1)$. Equivalently, you can first contract the graph of $y=f(x)$ by a factor of 2 to get the graph of $y=f(2x)$. You can then shift this left by $\frac{1}{2}$ to get the graph of $y=f\left(2\left(x+\frac{1}{2}\right) \right)=f(2x+1)$.$

Graphing Question (1 Viewer)

axwe7

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Shuuya

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Drsoccerball

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Green Yoda

Hi Φ

leehuan

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Shuuya

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axwe7

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KingOfActing

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Paradoxica

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Green Yoda

Hi Φ

KingOfActing

lukewarm mess

InteGrand

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