Okay /b/ help out a dirty spic nigger with his homework
Translate and I'll do it for you
>>604312907
What part don't you understand?
>>604313376
I don't speak spic. Therefore, I don't know what it's asking you to do.
>>604313528
I was asking OP. Unless you're OP, in which case I'm confused as to why you've been assigned math homework in a language that you don't speak.
>>604313904
Disregard. I blame the beer.
>>604313123
Ok man i'll translate it
The first one is Determine the image of zero by the function of f
>>604313123
"funcion"
"imagen"
"composicion"
"cero"
"determina"
"simplifica"
Seriously, all the words are transparent
>>604314636
Yeah it's a bilenguial school
>>604314401
1. -2
2. -1
>>604312907
I just got doing problems related to this.
#1 is asking you to graph f(x)=3x^2-2, if x=0 ("cero" means 0 in spanish)
#2 is asking you to graph g(x), if x=1
Want more help? Roll dubs faggot.
Oh gods of doubles give me the power
>>604314401
As long as "image" is translated correctly, this is asking what happens when you put zero into the function.
Think of a function as a machine. You put numbers into the input, and the machine spits out a different number. A rule like
f(x) = x^2
tells you how the function works. This rule for the function f says if you put in, say, three, then you will get out three squared, which is nine:
f(3) = 3^2 = 9
To figure that out, I just put in the 3 for my x to determine my output. Here's another example:
g(x) = 4 x - 9
g(2) = 4*2 - 9 = -1
So for this function, if you put in 2, you get out -1. The image of 2 for the function g is -1.
Now try problem 1: tell me what the image of 0 is for f. If you put 0 into f, what do you get out?
Do your homework yourself nigger
>>604312907
Where the fuck are you doing this homework? You're not in an english speaking country are you?
I've only heard of bilingual schools in latin american countries, and this material is basic so you're doing middle school grade shit.
>>604314960
No it's not, you faggot. The image of a function evaluated at a point is the corresponding member of the function's co-domain.
>>604314940
This guy's got it. To find the image, simply plug in the number it gives you for x.
For 3, simply plug in (x+h) for x. You'll get (x+h)^3+(x+h), then use binomial theorem (or wolframalpha) to expand.
For the fourth, evaluate the function at x = -1 for a, then plug in a+b for b), then plug in x-h for c).
For number five, it's asking you to solve 5 = x^2+x-1, which simplifies to 0 = x^2 + x - 6, which factors to (x-2)(x+3), so both -3 and 2 work for values of x.
The next part is just telling you the difference between an even and an odd function.
For the next part, you're just composing functions. For the first one f(circle)g, just plug the entire g(x) function into the f(x) function. That's all it is. Then do it backwards, plug the f(x) function into the g(x) function.
Hope that helps you mathematically illiterate faggot.
>>604316164
Puerto Rico man and for being a faggot i failed highschool
>>604316384
Thanks /b/ro yeah i know i'm a dumb faggot
damn, you're one dumb motherfucker.. Pay attention in class nigger
I have faith in u op
>>604316384
Shit, don't tell him the answers. Help him learn. If you think that post did any good, you must be teaching in a university.
>>604316384
>No it's not, you faggot.
1. Determina la imagen de cero bajo la funcion...
Translation: Determine the image of zero under the function...
>Determine the image
As in graph the thing.
Do you even spanish?
And he'll get the same answer as
>>604314940
From what I wrote, he would just have to read the graph.
>The image of a function evaluated at a point is the corresponding member of the function's co-domain.
You really think OP is going to understand that? kek
>>604312907
Still waiting for them dubs OP.
>>604318000
>Determine the image
>As in graph the thing.
Wrong. See the basic definition of a function, for example here:
http://www.cs.odu.edu/~toida/nerzic/content/function/definitions.html
The image of an element of the domain is the element of the co-domain to which it is paired in the relation. Given a general function f, the image of 0 is f(0). You can't even graph an image since an image is a set.
>>604312907
4. if f(x)=(x^2 -1)/x
(a) 0
(b) f(a+b)=[(a+b)^2 -1]/(a+b)
(c) f(x-h)=[(x-h)^2 -1]/(x-h)
>>604319029
Sure you can, it's a point at the corresponding coordinates on the graph of the equation. It doesn't ask for that, but you can graph the equation and just highlight the point for value 0.
Why so butthurt about this anon?