1. ## Domain and Range

i dont really know how to do them. i dont know how to do the squared sign so ill put a ! in front of the 2 if i need to put em.

1) y=x!2 so this one would be y equals x squared
2) y=x!2+3
3) y=(x+3)!2
4) y=3x!2
5) y=-x!2
6) y=1/2x!2
7) y=1/3(x-2)!2
8) y=-(x+2)!2+4

2. Originally Posted by bosstycoon510
i dont really know how to do them. i dont know how to do the squared sign so ill put a ! in front of the 2 if i need to put em.

1) y=x!2 so this one would be y equals x squared
2) y=x!2+3
3) y=(x+3)!2
4) y=3x!2
5) y=-x!2
6) y=1/2x!2
7) y=1/3(x-2)!2
8) y=-(x+2)!2+4

Hello, bosstycoon,

I assume from the title of your post that you want to know the domain and the range of the given functions:

The domain of a function is the set of real numbers which are "allowed" with this function. The range contains all possible numbers of y.

to 1) no restrictions for x, so $D=\mathbb{R}$
all squares are positve or zero.
therefore $y\geq0\Longrightarrow R=\mathbb{R_\text{0}^+}$

to 2)no restrictions for x, so $D=\mathbb{R}$
all squares are positve or zero. Now you add 3 to a pos. number or zero.
therefore $y\geq 3$

to 3) same as 1)

to 4) same as 1)

to 5) same domain as in 1).
But now you multiply the squares by (-1) so the domain contains only negative numbers or zero: $R=\mathbb{R_\text{0}^-}$

to 6) same as 1)

to 7) same as 1)

to 8) same domain as in 1).
But now you subtract numbers from 4.
Therefore the range contains all numbers which are smaller or equal 4: $y\leq 4$

Greetings

EB

3. so all of the domains are x is greater than or equal to zero?
srry neva seen the range written like that for 1) and 5) in my class b4 so im not sure of the ranges for em
and the ranges are:
1) y is greater than or equal to zero... i think
2) y is greater than or equal to three
5) i dont know what the range is becuz of the way u typed it
8) y is less than or equal to 4

4. Let me show you my secret of how to find ranges without graphs.

Let me do the last one.
$y=-(x+2)^2+4$
What you are going to do now is find all the possible $y$ such as the equation above has a real solution (no imaginary).
Subtract 4 from both sides,
$y-4=-(x+2)^2$
Multiply by -1,
$4-y=(x+2)^2$
A solution (real) exists only when the right want is non-negative thus,
$4-y\geq 0$ thus, $y\leq 4$

5. i still dont really get it. can u make up another problem and let me c how u do it cuz the one u showed me is weird how did u make
4-y=(x+2)^2 go to 4-y is greater than or equal to 0

6. Originally Posted by bosstycoon510
i still dont really get it. can u make up another problem and let me c how u do it cuz the one u showed me is weird how did u make
4-y=(x+2)^2 go to 4-y is greater than or equal to 0
I give you another problem, but let me explain this one. For example, if you have $x^2=n$ for what values of $n$ does the equation have a solution. Note, that if $n<0$ then you have $x^2$ being equal to a negative number which is impossible because $x^2\geq 0$. (You cannot square a number and get a negative).

Try this one, find the range of,
$y=\frac{1}{1+x^2}$

7. y is less than or equal to one?

8. Originally Posted by bosstycoon510
y is less than or equal to one?
Can y be negative?

RonL

9. darn i thought i had it too

10. You have,
$y=\frac{1}{1+x^2}$
Note first, $y\not = 0$ otherwise no solution cuz (no number when dividing 1 gives zero).

Therefore, you can take reciprocal of both sides since $y\not = 0$,
$\frac{1}{y}=1+x^2$
Thus,
$\frac{1}{y}-1=x^2$
As explained before this has a solution whenever,
$\frac{1}{y}-1\geq 0$
Thus,
$\frac{1}{y}\geq 1$
From here we see that $y$ is positive cuz otherwise the fraction is negative an no way can be larger than 1. Thus, $y>0$, take reciprocal of both sides (and the sign flips) thus,
$y\leq 1$

Thus,
$0