## Barrelracing one year ago When looking at the rational function f of x equals the quantity x minus one times the quantity x plus two times the quantity x plus four all divided by the quantity x plus one times the quantity x minus two times the quantity x minus four, Bella and Edward have two different thoughts. Bella says that the function is defined at x = –1, x = 2, and x = 4. Edward says that the function is undefined at those x values. Who is correct? Justify your reasoning.

1. Barrelracing

@jim_thompson5910

2. Barrelracing

@freckles

3. jim_thompson5910

 f of x equals the quantity x minus one times the quantity x plus two times the quantity x plus four all divided by the quantity x plus one times the quantity x minus two times the quantity x minus four what a mess... hopefully I translated the word problem into the proper function $\Large f(x) = \frac{(x-1)(x+2)(x+4)}{(x+1)(x-2)(x-4)}$ let me know if I did or not

4. Barrelracing

yes you did

5. jim_thompson5910

ok great

6. jim_thompson5910

Bella says that the function is defined at x = –1, x = 2, and x = 4 let's check this claim. If we were to replace EVERY x with a value like x = 4, then... $\Large f(x) = \frac{(x-1)(x+2)(x+4)}{(x+1)(x-2)(x-4)}$ $\Large f(4) = \frac{(4-1)(4+2)(4+4)}{(4+1)(4-2)(4-4)}$ $\Large f(4) = \frac{(3)(6)(8)}{(5)(2)(0)}$ $\Large f(4) = \frac{144}{0}$ $\Large f(4) = \text{Undefined}$ the result is undefined because we cannot divide by zero. We can't have 0 in the denominator

7. jim_thompson5910

Bella says that the function is defined at x = –1, x = 2, and x = 4 so Bella is wrong. At least about the x=4 part. I recommend you check x=-1 and x=2

8. Barrelracing

ok thank you

9. jim_thompson5910

no problem

10. Barrelracing

the rest are not undefined

11. jim_thompson5910

correct,x=-1 and x=2 are also undefined because they make the denominator 0

12. jim_thompson5910

oh wait, you should say "the rest are undefined" or "the rest are not defined"

13. Barrelracing

i got f(-1)=-4 and f(2)= 11

14. jim_thompson5910

you did something wrong

15. Barrelracing

f(-1) = (-1 - 1)(-1 + 2)(-1 + 4)/ (-1 + 1)(-1 - 2)(-1 - 4) = -2 + 1 + 3/ 0 - 3 - 5 = 2/-8 f(-1) = -4

16. Barrelracing

f(2) = (2 - 1)(2 + 2)(2 + 4)/ (2 + 1)(2 - 2)(2 - 4) = 1 + 4 + 6/ 3 + 0 - 2 = 11/1 = 11

17. Barrelracing

are you gonna help me still? @jim_thompson5910

18. jim_thompson5910

sorry my notification thing isn't working properly

19. Barrelracing

thats ok

20. jim_thompson5910

(-1 + 1)(-1 - 2)(-1 - 4) = 0*(-3)*(-5) = 0 is in the denominator

21. jim_thompson5910

the parenthesis next to each other means multiply

22. jim_thompson5910

make sense?

23. Barrelracing

ohhh ok thank you so much for the help.

24. jim_thompson5910

no problem