Question

Meddle and Fan
Describe the transformations of the functions
Picture attached

Answer 1

well the second function will move down vertically by 3

Answer 2

hence why it says - 3

Answer 3

From the parent function \(\Large \sf y=f(x)\)

Transformations:
\(\sf \LARGE y= \color{blue}{a}f(\color{red}{k}(x-\color{green}{d}))+\color{orange}{c}\)

\(\color{blue}{a}\) is vertical stretch/compression

|\(\color{blue}{a}\)| > 1 stretches
|\(\color{blue}{a}\)| < 1 compresses
\(\color{blue}{a}\) < 0 flips the graph upside down

\(\color{red}{k}\) is horizontal stretch/compression

|\(\color{red}{k}\)| > 1 compresses
|\(\color{red}{k}\)| < 1 stretches
\(\color{red}{k}\) < 0 flips the graph left-right

\(\color{green}{d}\) is horizontal shift

\(\color{green}{d}\) < 0 shifts to the right
\(\color{green}{d}\) > 0 shifts to the left

\(\color{orange}{c}\) is vertical shift

\(\color{orange}{c}\) > 0 shifts upward
\(\color{orange}{c}\) < 0 shifts downward


Your parent function is \(\Large \sf y=1/x\)
Apply Transformations:
\(\sf \LARGE y=\frac{ \color{blue}{a}}{(\color{red}{k}(x-\color{green}{d}))}+\color{orange}{c}\)

Answer 4

so for the first 1.

\(\color{blue}{a}\) = -2, less than 0, so the function will ? it is also greater than 1, so it will?
\(\color{green}{d}\)= 1 , greater than 0, so it will ?

Answer 5

for the second, you have three transformations that you have to apply not only the vertical shift

Answer 6

flips the graph upside down
and shifts to the right?

Answer 7

oh okay

Answer 8

for the first one, yes you're right flip upside down and ..? (vertical stretch or compression)
and yes it will shift horizontally right (:

Answer 9

compress

Answer 10

nope, \(\color{blue}a\) is GREATER than 1.. so it should be?

Answer 11

stretch

Answer 12

yes (:
can you do the second one?

Answer 13

i think so :)

Answer 14

try it, ill check it for you

Answer 15

vertical stretch, horizontal shift to the right, and vertical shift down?

Answer 16

right, you got it! ^_^

Answer 17

YAY!!! Thank you so much!

Answer 18

no problem, as long as you learn ;p

goodluck!