Question

Will somebody please help me with this Algebra??

Answer 1

Given the function f(x)=4(x+1)^2-3, indicate the shifts that will effect the location of the vertex and explain what effect they will have. Use complete sentences.
> f(x-2)
> f(x)-2
> f(2x)
> 2*f(x)

Answer 2

@bbbbbbbbbbbbbbbbbb ??

Answer 3

f(x-2) moves the vertex two units to the right.
f(x) -2 moves the vertex two units down
..

Answer 4

I got that for the first one.
Thanks for the second :)
I don't know how to do the rest, though. :/

Answer 5

ok so lets think about the third for example
f(2x)

now what was in x=2 before the transformation is in x=1 after the transformation right ?
can you tell me what it does then ?

Answer 6

What? No, I don't get it.
Tbh, I just guessed on how to do the first one.
I don't have any real clue how to do it all. :/

Answer 7

look:
f(x)
when we plug x =1 we get f(1)
when we plug x=2 we get f(2)


f(2x)
when we plug x = 0.5 we get f(1)
when we plug x=1 we get f(2)

Answer 8

can you see the pattern ?

Answer 9

yeah I see the pattern!

Answer 10

so we can understand that the x point of the vertex is now 1/2 of the x point of the vertex before the trans!

if it was -1 before now it is -0.5

Answer 11

the 1/2 is because of the 2 of curse.

Answer 12

what if we had f(3x) ?

Answer 13

Oh Lord... um.. 1.5?

Answer 14

what 1.5 ?

Answer 15

if we change
f(x)
to f(3x)

what value of x do we need to plug into f(3x) to get f(x) again ?

Answer 16

I have no idea

Answer 17

look, before the trans our function was
f(x)

if we plug into the function x=b
we get
f(b)

now after the trans we have
f(3x)

in order to get the same f(b)
we need to plug b/3

Answer 18

because f(3*b/3) = f(b)

Answer 19

Oh okay! I would never have gotten that.. Lol

Answer 20

same for
f(x)
and
f(x-2)

before the trans when x=b
we get
f(b)

after the trans we need to plug
x=b+2
in order to get
f(b) again
because f(b+2-2)=f(b)
thus we moved the point 2 unites to the right because x=b turned to x=b+2

Answer 21

\(\large \begin{array}{rllll}
f(x)=4(x+1)^2-3\\ \quad \\
f(\color{red}{x-2})\implies &4[(\color{red}{x-2})+1]^2-3\\ \quad \\
f(x)\color{red}{-2}\implies &[4(x+1)^2-3]\color{red}{-2}\\ \quad \\
f(\color{red}{2x})\implies &4[(\color{red}{2x})+1]^2-3\\ \quad \\
\color{red}{2}\cdot f(x)\implies &\color{red}{2}\cdot [4(x+1)^2-3]
\end{array}\)