# Thread: Domain of Function;Graphing;Zeros algebraically

1. ## Domain of Function;Graphing;Zeros algebraically

I have four problems I need help with.

The first two are-
Find the domain of the function:
√25-x2
Edit: After attempting this problem,I've gotten (-5
,5) ..is that right?

and
Find the domain of the function:
x/x2-x-6
Edit: For this one I've gotten (-∞,-2)U(-2,3)U(3,∞).Once again,I'd appreciated some double checking from someone.

The next is-
Find the zeros of the function algebraically:
f(x)=x3-x2-25x+25
Edit: And for this one,I've gotten 5,-5,1.Correct?

The last is-
Graph the function:
x2-2, x<-2
f(x)= 5 , -2< x <0

8x-5, x >0

^This one specifically I really need help understanding. ):

2. ## Re: Domain of Function;Graphing;Zeros algebraically

For your first check, the √25-x2 is not limited in any domainas it is a parabola opening downward. If graphing is not available simply look at the equation... is there any value of x that would not allow it to be a real number? because x can only be squared, there are no limits.

Again, with your second problem there are no limits on what x can be (which is what a domain is) except that it cannot be zero because if it was zero, then the fraction would become undefined. That would get you a domain of all reals excluding zero because zero squared would equal zero.

Your third is correct

This last one is called a piecewise function, where each segment if the function has the properties of a different toolkit. so, to help you understand, i will write out how that equation would sound. "y will equal X squared minus 2 when X is less than negative 2". that is the first part. So wherever x is less than -2, you would graph x^2-2, and then stop at -2 with an open circle because it cannot actually equal -2. The next would read like "F of X equals 5 (Y=5) when x is greater than or equal to -2 and less than of equal to 0" same deal. Finally, " will equal 8x minus 5 when x is greater than zero (not including zero)" to symbolize an end that includes the final value is a closed circle.

Hope this has helped

3. ## Re: Domain of Function;Graphing;Zeros algebraically

My understanding was that the restriction for the first problem is anything that results in a negative root,because a negative root isn't possible (without getting into imaginary numbers).And for the second,my reasoning was that when -2 is plugged into the denominator you get 4+2-6,which would be zero and when you plug in 3 you get 9-3-6,which again is a zero;while anything below -2,above 3,and inbetween excluding those numbers would result in either a negative denominator or a positive one,hence my interval notation.
I believe your last explanation does help me understand better (especially where the closed/open circles are concerned) and I'm trying very hard to understand,so thank you for replying back!

4. ## Re: Domain of Function;Graphing;Zeros algebraically

Originally Posted by Ai Ekio
My understanding was that the restriction for the first problem is anything that results in a negative root,because a negative root isn't possible (without getting into imaginary numbers).And for the second,my reasoning was that when -2 is plugged into the denominator you get 4+2-6,which would be zero and when you plug in 3 you get 9-3-6,which again is a zero;while anything below -2,above 3,and inbetween excluding those numbers would result in either a negative denominator or a positive one,hence my interval notation. I believe your last explanation does help me understand better (especially where the closed/open circles are concerned) and I'm trying very hard to understand,so thank you for replying back!
I think that FxAperture read the question as $\displaystyle \sqrt{25}-x^2$ and not as $\displaystyle \sqrt{25-x^2}$.
That is because you failed to use any grouping symbols at all in the entire post.
For example x/x^2-x-6 should be read as $\displaystyle \frac{x}{x^2}-x-6$, but you meant $\displaystyle \frac{x}{x^2-x-6}$. So group things.

5. ## Re: Domain of Function;Graphing;Zeros algebraically

Originally Posted by Plato
I think that FxAperture read the question as $\displaystyle \sqrt{25}-x^2$ and not as $\displaystyle \sqrt{25-x^2}$.
That is because you failed to use any grouping symbols at all in the entire post.
For example x/x^2-x-6 should be read as $\displaystyle \frac{x}{x^2}-x-6$, but you meant $\displaystyle \frac{x}{x^2-x-6}$. So group things.
Oh,okay.Sorry,I'm not very literate when it comes to how to type out equations.Thank you.

6. ## Re: Domain of Function;Graphing;Zeros algebraically

Hello, Ai Ekio!

FxAperture did an excellent job explaining the piecewise function.
I'll baby-step through its graphing . . .

$\displaystyle \text{Graph the function: }\:f(x) \;=\;\begin{Bmatrix}x^2-2 && x < -2 \\ 5 && -2 \le x \le 0 \\ 8x-5 && x > 0 \end{array}$

The first function is a parabola: ,$\displaystyle y \:=\:x^2-2$
Code:
              |
*        |        *
|
|
|
*       |       *
|
|
*      |      *
|
*     |     *
-----*----+--------
*  |  *
*
|
But we want only the portion less than -2.
Code:
              |
*        |          .
|
|
|
*       |       .
|
|
o      |      .
:      |
:.     |     .
---+-.----+----.---
-2   .  |  .
.
|

|
o|
|

The second function is: .$\displaystyle y \:=\:5$, a horizontal line.
Code:
              |
|
* * * * * * * * *
|
|
|
|
----------+--------
|
|
But we want the portion between -2 and 0, inclusive.
Code:
              |
|
********
:      |
:      |
:      |
:      |
---+------+--------
-2      |
|

The third function is a line.
Code:
              |
|      *
|
|     *
|
|    *
|
|   *
|
|  *
|
----------+-*------
|
|*
|
*
|
*|
|
But we want the portion greater than zero.
Code:
              |
|      *
|
|     *
|
|    *
|
|   *
|
|  *
|
----------+-*------
|
|*
|
o
|
|
|

Therefore, the graph looks like this:

Code:
              |
*        |      *
|
|     *
|
*       |    *
********
|   *
o      |
:      |  *
:      |
---+------+-*------
-2      |
*
|
o
|

7. ## Re: Domain of Function;Graphing;Zeros algebraically

Originally Posted by Soroban
Hello, Ai Ekio!
Thank you!The visuals really helped. (: