## JacksonJRB one year ago Algebraically determine the domain and range:

1. JacksonJRB

Typing the equation now

2. SolomonZelman

3. JacksonJRB

|dw:1440467179872:dw|

4. JacksonJRB

Equation button isn't working

5. SolomonZelman

You can copy paste my code (it works in text) $$\large\color{black}{ \displaystyle y=\sqrt{x^2-4x-5} }$$

6. SolomonZelman

$$\large\color{black}{ \displaystyle y=\sqrt{x^2-4x-5} }$$

7. SolomonZelman

So, you know that the part inside the square root must not be negative, right?

8. JacksonJRB

Yes

9. JacksonJRB

There we go, had to restart my browser

10. SolomonZelman

So, set: $$\large\color{black}{ \displaystyle x^2-4x-5\ge0 }$$ and solve for x

11. SolomonZelman

this will give you the domain

12. JacksonJRB

What about the range? That's what really always confuses me

13. SolomonZelman

Well, $$\large\color{black}{ \displaystyle x^2-4x-5 }$$ will equal to 0 for at least 1 value of x, and if $$\large\color{black}{ \displaystyle x^2-4x-5=0}$$ for some x-value, then we know that y is =0 for some x-value. Now, you can probably tell answer these two questions without me:                                                 1) Will you ever get a negative output for y? -------------------------------------------------- 2) $$\large\color{black}{ \displaystyle x^2-4x-5 }$$ can get as large as you want, $$\\[0.8 em]$$ and thus $$\large\color{black}{ \displaystyle \sqrt{x^2-4x-5} }$$ can also get as large as you want. $$\\[0.9 em]$$ Knowing this, is there a limit for how large the range (the output) can be?

14. JacksonJRB

1) no 2) no

15. SolomonZelman

Yes, correct...

16. SolomonZelman

So can you tell me the range ?

17. SolomonZelman

(if you got questions, it's ok, ask)

18. JacksonJRB

|dw:1440467760532:dw|

19. SolomonZelman

Yes, that is right

20. JacksonJRB

Is that right?

21. SolomonZelman

yes

22. JacksonJRB

Ah ok

23. JacksonJRB

Thanks

24. SolomonZelman

Yes, you are welcome!

25. SolomonZelman

And the domain is?

26. JacksonJRB

|dw:1440467835822:dw|

27. SolomonZelman

$$\large\color{black}{ \displaystyle y=\sqrt{x^2-4x-5} }$$ You know that the part inside the square root can't be negative: So we set: $$\large\color{black}{ \displaystyle x^2-4x-5\ge0 }$$ $$\large\color{black}{ \displaystyle (x-5)(x+1)\ge0 }$$ For left side to be greater than 0, factors have to be either both negative or both positive or one of factors has to be equal to 0. $$\large\color{black}{ \displaystyle x\le-1 }$$ then for values less than -1, both factors are negative (and - times - =+ , and +>0) Or when x=-1 one factor is zero, and 0$$\ge$$0 which also satisfies the equation. $$\large\color{black}{ \displaystyle x\ge5 }$$ when x=5 we again get $$0\le0$$ (which satisfies the equation). For values greater than 5, both factors are positive (and + times + = +, which is greater than 0 as well)

28. JacksonJRB

Whoops, got the signs mixed

29. SolomonZelman

However if you said $$x\ge-1$$ (you made that little mistake) then any number between -1 and 5 will disprove the statement. For example x=0, $$\large\color{black}{ \displaystyle (x-5)(x+1)\ge0 }$$ $$\large\color{black}{ \displaystyle (0-5)(0+1)\ge0 }$$ $$\large\color{black}{ \displaystyle (-5)(1)\ge0 }$$ $$\large\color{black}{ \displaystyle -5\ge0 }$$ and that is not true.

30. SolomonZelman

So, as I showed, the solutions are $$x\le -1$$ $$x\ge 5$$

31. JacksonJRB

Thanks again!

32. SolomonZelman

Yw