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I Want to know if what I say about transforming (and shifting of f(x)) right or not? We want to draw $af(bx-c)+d$ from $f(x)$:

  1. $af(bx-c)+d = af(b(x-\frac{c}{b}))+d$.

  2. We shift f on x-axis by $c/b$

  3. We transform f on x axis so we multiply its Xs by 1/b.

  4. We transform f on on y axis so we multiply tis Ys by a.

  5. Finally we shift f on y-axis by d.

My question is specially about 2 and 3 to know if they are right or not? (specially about order of doing this) (I know the negative transformations, so think all of a,b,c,d all positive).

Thanks.

3 Answers 3

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If $c/b$ is positive, then subtracting it inside will move it to the right. Think about it this way:

$$f(x)\to f(x-2)$$

For it to be the same, $x$ must be $2$ units larger to cancel the $-2$.

Similarly, imagine the following:

$$f(x)\to f(2x)$$

For it to stay the same, $x$ must be half the original size, hence, we divide by $b$.

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    My exact question is that "Is it right to first shift $c/b$ and then transforming with $b$ (multiply by $1/b$))?" (My question is specially about order of doing this and not sending to right or left) – 2017-01-04
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    @titansarus Oh, the order of operations? Well, $b(x-c/b)$, so the inner most operation must be handled first. It is inside with the $x$, while the $b$ on the outside does not affect $x$ until after we subtract $c/b$. – 2017-01-04
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    So the order of what I said is True, yes? – 2017-01-04
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    @titansarus Yes, the order is correct. But number $5$ should be on the $y$-axis. – 2017-01-04
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    Thank you very much. It was spelling error. – 2017-01-04
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    @titansarus No problem. Small typos are unavoidable in day to day life. – 2017-01-04
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    @SimpleArt Can we creating $f(bx-c)$ by this way ? firstly create $f(bx)$ and then create $f(bx-c/b)$ – 2017-01-15
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    @S.H.W Sure we can. As you say, first we create $f(bx)$, but instead of ending up with $f(bx-c/b)$, let $x\to x-c/b$ so that we end up with $f(bx-c)$. I'm not sure how you ended up with $f(bx-c/b)$ there. – 2017-01-15
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    @SimpleArt Yes thank you , I made a mistake. But I've two problems. First , When we do a horizontal multiplication of $1/a$ we multiple only $x$ by $a$ or multiple $x+b$ by $a$. Second , in this site(http://mathforum.org/library/drmath/view/68503.html) said that : – 2017-01-15
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    "When we wrote the transformed function as $f(ax+b)$, the operations inside the function were done in the order "multiply, then add"; since we are replacing x in each transformation, we ended up doing the last operation first (outside in), replacing x with $x+b$ first, then with a, so we did the shift followed by the shrink. When we wrote it as $f(a(x+b/a))$, the order of operations says the addition comes first, then the multiplication, so the transformations have to be done in the opposite order." But I do same order as $f(ax+b)$. Is it wrong ? – 2017-01-15
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    @S.H.W Well, order of operations also says that we should do parenthesis first. So as long as you place your parenthesis so that it comes out right, you should be fine. – 2017-01-15
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    @SimpleArt Okay , but in $f(ax+b)$ we do "multiply , then add" according to reference we have to first add then multiply for drawing function – 2017-01-15
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    @S.H.W No, we can multiply then add. After multiplying, we just move the resulting graph $b$ units in whichever direction – 2017-01-15
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    @SimpleArt Okay , so this sentence isn't necessary "The transformations have to be done in the opposite order" . Am I right ? also can you answer to first question ? – 2017-01-15
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    @S.H.W Yes, you are right. It really depends on the parenthesis and what-not. To what I think is the first question: You multiply $x+b$. – 2017-01-15
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51843/discussion-between-s-h-w-and-simple-art). – 2017-01-15
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Yes, the order of what you wrote is correct.

For another perspective, one of the schools where I used to teach College Algebra taught the HSRV method for graphing transformations:

  1. H = horizontal shifts (this is the same as your step 2)
  2. S = stretches and shrinks, both horizontal and vertical (this is the same as your step 4)
  3. R = reflections, both over the $x$-axis and $y$-axis (this isn't explicit in your steps but it could be nestled in with your steps 3 and 4)
  4. V = vertical shifts (this is the same as your step 5)

A little different from the steps you list out, but the point is that horizontal shifting does come before the horizontal stretching/shrinking

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    Can we creating $f(bx-c)$ by this way ? firstly create $f(bx)$ and then create $f(bx-c/b)$ – 2017-01-15
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If you know what the graph of $y=f(x)$ looks like, you can obtain the graph of $$\frac{y-y_0}{B} = f\left(\frac{x-x_0}{A}\right)$$ immediately: just scale horizontally by a factor of $A$, scale vertically by a factor of $B$, then translate horizontally by $x_0$ and vertically by $y_0$.

How on earth to remember this? You probably already know the equation of an ellipse, and it uses exactly the same form: $\left(\frac{x-x_0}{A}\right)^2 + \left(\frac{y-y_0}{B}\right)^2 = 1$. You think of this as being obtained from the unit circle $x^2+y^2=1$ by exactly the same scaling and translations.

In your specific case, you have $$y=af(bx-c)+d$$ $$y-d = af(bx-c)$$ $$\frac{y-d}{a}=f\left(b(x-\tfrac cb)\right)$$ $$\frac{y-d}{a}=f\left(\frac{x-\tfrac cb}{\tfrac 1b}\right)$$

This is in the desired form with $x_0=c/b$, $y_0=d$, $A=a$, and $B=1/b$. So we can say that you just take the original graph of $y=f(x)$, scale it horizontally by a factor of $1/b$, scale it vertically by a factor of $a$, translate horizontally by $c/b$, and translate vertically by $d$. Done.

The exact same reasoning applies to equations of the more general form $$F\left(\frac{x-x_0}{A},\frac{y-y_0}{B}\right)=0$$ In fact, that's the ellipse I mentioned. In that case, $F(x,y)=x^2+y^2-1$.

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    Can we creating $f(bx-c)$ by this way ? firstly create $f(bx)$ and then create $f(bx-c/b)$ – 2017-01-15