Algebra 2

what the hell does this mean?g(x)=4(x-2)

“You’ll never know how strong you really are until being strong is the only choice you have left”

hello? anyone here
*echo*

“You’ll never know how strong you really are until being strong is the only choice you have left”

Abaddon angel of destruction wrote: what the hell does this mean?g(x)=4(x-2)

g(x) means the same as y
So that means y=4(x-2)

new problem... anyone care to help?

“You’ll never know how strong you really are until being strong is the only choice you have left”

Abaddon angel of destruction wrote: new problem... anyone care to help?

It’s equal to five.

-King- wrote:

Abaddon angel of destruction wrote: new problem... anyone care to help?

It’s equal to five.

i’m going to post a new one i mean

“You’ll never know how strong you really are until being strong is the only choice you have left”

graph function. label the vertex, axis of symmetry and x- intercepts

k here it goes:
y=-(x-3)(x+1)

so x is 3, -1 right?
and it opens down.

i’m lost after that, what do i do next?

“You’ll never know how strong you really are until being strong is the only choice you have left”

Abaddon angel of destruction wrote: graph function. label the vertex, axis of symmetry and x- intercepts

k here it goes:
y=-(x-3)(x+1)

so x is 3, -1 right?
and it opens down.

i’m lost after that, what do i do next?

bump, anyone care to help please?

“You’ll never know how strong you really are until being strong is the only choice you have left”

Abaddon angel of destruction wrote: graph function. label the vertex, axis of symmetry and x- intercepts

k here it goes:
y=-(x-3)(x+1)

so x is 3, -1 right?
and it opens down.

i’m lost after that, what do i do next?

X-intercepts would be (3,0) and (-1,0).

Vertex would be...

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

-b/2a=x coordinate of vertex
(-2)/2(-1)=1

Vertex: (1, 4)


Axis of Symmetry: X=1

MvC Chat needz moar people
http://xat.com/chat/room/60567900/

Marly wrote:

Abaddon angel of destruction wrote: graph function. label the vertex, axis of symmetry and x- intercepts

k here it goes:
y=-(x-3)(x+1)

so x is 3, -1 right?
and it opens down.

i’m lost after that, what do i do next?

X-intercepts would be (3,0) and (-1,0).

Vertex would be...

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

-b/2a=x coordinate of vertex
(-2)/2(-1)=1

Vertex: (1, 4)


Axis of Symmetry: X=1

english please...? wait never mind i think i got how you did it, ok thanks

“You’ll never know how strong you really are until being strong is the only choice you have left”

i’m stuck on this one, i dont know what i did wrong. help please?

y=3(x-1)(x-4)

x=1,0 and 4,0

standard form:y=3x^2-15x+12

i know its wrong cuz its supposed to go up and i got a negative, i got:
for the vertex:

2 1/2, -6 3/4

any help?

“You’ll never know how strong you really are until being strong is the only choice you have left”

Abaddon angel of destruction wrote: i’m stuck on this one, i dont know what i did wrong. help please?

y=3(x-1)(x-4)

x=1,0 and 4,0

standard form:y=3x^2-15x+12

i know its wrong cuz its supposed to go up and i got a negative, i got:
for the vertex:

2 1/2, -6 3/4

any help?

It does go up. Your vertex is below the X-axis, and the X-intercepts are above your vertex. You did it correctly.

MvC Chat needz moar people
http://xat.com/chat/room/60567900/

Marly wrote:

Abaddon angel of destruction wrote: i’m stuck on this one, i dont know what i did wrong. help please?

y=3(x-1)(x-4)

x=1,0 and 4,0

standard form:y=3x^2-15x+12

i know its wrong cuz its supposed to go up and i got a negative, i got:
for the vertex:

2 1/2, -6 3/4

any help?

It does go up. Your vertex is below the X-axis, and the X-intercepts are above your vertex. You did it correctly.

oh... yeah haha im an idiot i see it thanks. i just tried to graph it as positive six for some reason

“You’ll never know how strong you really are until being strong is the only choice you have left”

k now i’m stuck on a review problem that i forgot how to do: y=2/x/

^those thingies are supposed to be absolute value, and its x intercept

“You’ll never know how strong you really are until being strong is the only choice you have left”

Abaddon angel of destruction wrote: k now i’m stuck on a review problem that i forgot how to do: y=2/x/

^those thingies are supposed to be absolute value, and its x intercept

please anyone i have a benchmark tomorrow

“You’ll never know how strong you really are until being strong is the only choice you have left”

ugh... where are ty and marly when i need them...

“You’ll never know how strong you really are until being strong is the only choice you have left”

Abaddon angel of destruction wrote: k now i’m stuck on a review problem that i forgot how to do: y=2/x/

^those thingies are supposed to be absolute value, and its x intercept

X intercept of y=2|x|? Just do the math like any equation, y=0
so 0=2|x|
In this case you don’t have to worry about the absolute value sign for now, but later on you’ll have things like y=2|x-6|-2. You’ll just have to know how the graph changes with the 2, -6, and -2 added in from the normal y=|x|, but that’s later on.

Tyreaus Dreacon wrote:

Abaddon angel of destruction wrote: k now i’m stuck on a review problem that i forgot how to do: y=2/x/

^those thingies are supposed to be absolute value, and its x intercept

X intercept of y=2|x|? Just do the math like any equation, y=0
so 0=2|x|
In this case you don’t have to worry about the absolute value sign for now, but later on you’ll have things like y=2|x-6|-2. You’ll just have to know how the graph changes with the 2, -6, and -2 added in from the normal y=|x|, but that’s later on.

so you dont turn both of them zero?

“You’ll never know how strong you really are until being strong is the only choice you have left”

Abaddon angel of destruction wrote:

Tyreaus Dreacon wrote:

Abaddon angel of destruction wrote: k now i’m stuck on a review problem that i forgot how to do: y=2/x/

^those thingies are supposed to be absolute value, and its x intercept

X intercept of y=2|x|? Just do the math like any equation, y=0
so 0=2|x|
In this case you don’t have to worry about the absolute value sign for now, but later on you’ll have things like y=2|x-6|-2. You’ll just have to know how the graph changes with the 2, -6, and -2 added in from the normal y=|x|, but that’s later on.

so you dont turn both of them zero?

If it’s just your x-intercept no, just the Y.

Tyreaus Dreacon wrote:

Abaddon angel of destruction wrote: k now i’m stuck on a review problem that i forgot how to do: y=2/x/

^those thingies are supposed to be absolute value, and its x intercept

X intercept of y=2|x|? Just do the math like any equation, y=0
so 0=2|x|
In this case you don’t have to worry about the absolute value sign for now, but later on you’ll have things like y=2|x-6|-2. You’ll just have to know how the graph changes with the 2, -6, and -2 added in from the normal y=|x|, but that’s later on.

so the x intercept is zero too? wait i’m lost can you explain in a simpler form?

“You’ll never know how strong you really are until being strong is the only choice you have left”