Binomial Raised to a Power in a Line Equation?

[Iceycold12]

Hello. My teacher didn't teach how to distribute & powers in line equations. this but a friend helped me out with it but still can't figure how he got the solution.

I'm asked to find the translation of the graph between

and

. I get answer choices like this: up 4 and 5 to the right etc..

For

he somehow arrived at

. And said the vertex is at

.

Am I supposed to distribute the powers to everything inside the parenthesis and combine like terms?

If I gave a shot at it:

1. Distribute ^2 to everybody, inside parenthesis
2.

3. Combine Like Terms?
4.

Thanks.

 

[Deleted member 4993]

Hello. My teacher didn't teach how to distribute & powers in line equations. this but a friend helped me out with it but still can't figure how he got the solution.

I'm asked to find the translation of the graph between

and

. I get answer choices like this: up 4 and 5 to the right etc..

For

he somehow arrived at

. And said the vertex is at

.

Am I supposed to distribute the powers to everything inside the parenthesis and combine like terms?

Thanks.

I do not see why do you think - you'd need that kind of procedure or operation (and to my knowledge no such procedure exist).

Going back to your teacher's example:

y = (x+3)² - 2 .... add 2 to both sides of equation to get

y + 2 = (x+3)²

It is a well-known property of parabola - that states - if the equation of the parabola is

y - k = (x - h)²

Then the vertex of the parabola is at (h,k)

I'm asked to find the translation of the graph between

and

The easiest thing for you would be to use your graphing calculator - plot those two functions on same screen and observe....

 

[Iceycold12]

I see, would the same procedure apply to slope intercept lines? \(\displaystyle y=mx+b\)?

edit: Nevermind, answered myself, slope intercept for \(\displaystyle y=mx+b\) has no vertex.