1. Transformations

$f(x) = x^4 - x$

$g(x) = 16x^4 + 2x +1$

List the sequence of transformations that is necessary to obtain the graph of $g(x)$ from $f(x)$

How to do this question algebraically using $x'$ and $y'$ method.

2. Hello, usagi_killer!

$\displaystyle f(x) \:=\:x^4-x$
$\displaystyle [g(x) \:=\:16x^4 + 2x + 1$

List the sequence of transformations that is necessary to obtain the graph of $\displaystyle g(x)$ from $\displaystyle f(x).$

How to do this question algebraically using $\displaystyle x'$ and $\displaystyle y;$ method?
I "eyeballed" the two functions . . .

Start with: .$\displaystyle f(x)\:=\:x^4 - x$

Replace $\displaystyle x$ with $\displaystyle \text{-}2x\!:\;\;f(\text{-}2x) \;=\;(\text{-}2x)^4 - (\text{-}2x) \;=\;16x^4 + 2x$

Add 1: .$\displaystyle f(\text{-}2x) + 1 \;=\;16x^4 + 2x + 1$

. . Hence: .$\displaystyle g(x) \;=\;f(\text{-}2x) + 1$

First, graph $\displaystyle f(x).$

The sequence of transformations is:

(1) Replace $\displaystyle x$ with $\displaystyle -x$.
. . .This reflects the graph of $\displaystyle f(x)$ over the $\displaystyle y$-axis.

(2) Replace $\displaystyle x$ with $\displaystyle 2x.$
. . .This contracts the graph to "half its width."

. . .This raises the graph one unit.

The result is the graph of $\displaystyle g(x).$

3. Thank you!!!

But is there no way of doing it with mapping method?

Eg, if you had the graph y = x^2 and y = (x-1)^2

then let (x',y') be the new x and y

so y' = (x'-1)^2

thus y = y' and x = x'-1

so x' = x+1

thus (x,y) is mapped onto (x+1,y)

so the graph is moved 1 unit to the right.

How can you do the same way here?