Graphing $af(bx-c)+d$?

Question

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.

Answer 1

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 $b$.

Answer 2

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

Answer 3

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, $F(x,y)=x^2+y^2-1$.