List the sequence of transformations that is necessary to obtain the graph of from
How to do this question algebraically using and method.
Hello, usagi_killer!
I "eyeballed" the two functions . . .$\displaystyle f(x) \:=\:x^4-x$
$\displaystyle [g(x) \:=\:16x^4 + 2x + 1$
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?
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."
(3) Add 1.
. . .This raises the graph one unit.
The result is the graph of $\displaystyle g(x).$
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?