Question

given f(x)=x^2 sketch the graph and state the domain and range f(x)=-1/2[2(x+3)^2]-2

Answer 1

is it true that f(x)=-1/2[2(x+3)^2]-2 can be obtained from f(x)=x^2 by vertically compressing it by a factor of 1/2, horizontally compressing by 3, reflection on the x-axis, horizontal translation to the left by 3, and vertical translation up by 2?

Answer 2

(x-h)^2 +a
gives a right shift by h and an up shift by a.

Answer 3

soo...was what i said right though??

Answer 4

Take a look: http://www.wolframalpha.com/input/?i=plot [y+%3D+x^2%2C+y%3D+-1%2F2%282%28x%2B3%29^2%29-2%2C+{x%2C-10%2C10}%2C{y%2C-10%2C10}]
(or paste
plot[y = x^2, y= -1/2(2(x+3)^2)-2, {x,-10,10},{y,-10,10}]
into the Wolframalpha box)

Answer 5

is y= -1/2(2(x+3)^2)-2 the same as f(x)=-1/2(2x+6)^2-2?

Answer 6

Yes ( I was wondering about that, it seemed strange to put -1/2 * 2, it's just -1)

Answer 7

so wait do i just keep it as -1(2(x+3)^2)-2 then?

Answer 8

You can't distribute the 2 inside the parenthesis because of order of operations. Exponents first, then multiplication.

Answer 9

y= -1/2(2(x+3)^2)-2 is same as y =-(x+3)^2-2

Answer 10

"You can't distribute the 2 inside the parenthesis"

Sure I can, a(bc) = ab(c)

Answer 11

how come when i graph f(x) =-1/2(2x+6)^2-2, i don't get the same graph as -1/2(2(x+3)^2)-2?

Answer 12

You have put 2*2 inside instead of 2 (it's squared).

(Sorry, I see that i said it was the same)

Answer 13

But the answer I gave you is correct.

Answer 14

@smoothmath what if it's f(x)=-1/2(2x+6)^2-2? could i take the 2 out of the parentheses?

Answer 15

Estudiar, I was pointing out this:
2(x+3)^2 does not give
(2x+6)^2

Answer 16

Because alexeis_nicole asked
"is y= -1/2(2(x+3)^2)-2 the same as f(x)=-1/2(2x+6)^2-2?"

Answer 17

Oh, and no. You couldn't, nicole.

Answer 18

Yes, sorry, I thought you were talking about the associative operation.

Answer 19

=)

Answer 20

sooooo i just need to keep it as the original equation?

Answer 21

y =-(x+3)^2-2

Answer 22

ie you can cancel the -1/2 with the 2 like I said before

Answer 23

I agree.

Answer 24

what about f(x)=-1/2(2x+6)^2-2?
-1/2 still cancels out 2?

Answer 25

You already asked this, that is a different equation.

Answer 26

IF the problem was
-1/2(2x+6)^2-2? (The original problem WAS NOT)
then no, you could not multiply (-1/2) and 2.

Answer 27

I agree; and we have gone a bit far from the original question now.

Answer 28

okay, lol. well that means that the domain is a set of real number and the range would be y ≤ -2

Answer 29

That is yet another different question (post a new one, close this one)