## anonymous one year ago Find the vertical asymptotes, if any, of the graph of the rational function. Show your work. f(x) = x-4/x(x-4)

1. anonymous

x=1 is your vertical asymptote though if there was a factor of x-1 on top then it could possibly be a hole that is f(x)=(x-1)/(x-1) has a hole at x=1 (not a vertical asymptote) that is f(x)=x(x-1)/(x-1) also has a hole at x=1 but g(x)=(x-1)/(x-1)^2 has a vertical asympote at x=1 since it has more (x-1)'s on bottom then on top my answer Teacers reply>You find the VA's by setting the denominator to 0. When we get a factor that cancels out, it is not a vertical asymptote. What is it? Graph it to see what it looks like

2. anonymous

@campbell_st

3. anonymous

Is this from a test or something?

4. anonymous

no its from my work

5. anonymous

i havent taken my midterm yet lol

6. Jack1

f(x) = x-4/x(x-4) is that: $$\Large f(x) = x- \frac4x \times(x-4)$$ or $$\Large f(x) = x- \frac4{x \times(x-4)}$$ or $$\Large f(x) = \frac{x-4}{x \times(x-4)}$$

7. anonymous

|dw:1439383403225:dw|

8. Jack1

ok... so find when we're dividing by zero so denominator = x(x-4) ... when does this = 0?

9. anonymous

when we divide bye zero

10. Jack1

a(b) = 0 so either a = 0 or b = 0 in this case...a = x... so if a = 0, x = 0 a(b) = 0 in this case... b = x-4 if b = 0... x-4 = 0 ... so x = 4 vertical asymtotes at x=0 and x=4... yeah?

11. Jack1

is that how it works?

12. Jack1

or do u have to simplify the equation first?

13. anonymous

idk i gave my answer to the teacher and she said i was wrong so i dont know what to do

14. Jack1

kk $$\huge f(x) = \frac{x-4}{x \times(x-4)}$$ $$\huge f(x) = \frac{\color{red}{x-4}}{x \times\color{red}{(x-4)}}$$ $$\huge f(x) = \frac{1}{x}$$

15. Jack1

so if that's ur eqn.. then simplified it's f(x) = 1/x so when denominator = 0, then u have vertical asymptotoes only 1 v.a in this equation, it's when x = 0

16. Jack1

right...?

17. anonymous

right

18. Jack1

k... k bye