Question

I need help getting started with this question!
When looking at a rational function, 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. Describe a situation where Bella is correct, and describe a situation where Edward is correct. Is it possible for a situation to exist that they are both correct? Justify your reasoning.

Answer 1

hmmm are you covering asysmtotes?

Answer 2

or I should say, have you covered asymptotes yet, is this related to the section on asymptotes

have you covered quadratics yet?

Answer 3

@jdoe0001 Yes, I have covered all those things.

Answer 4

if we had a quadratic, whose root is say -> x = −1, x = 2, and x = 4 <-

do you know how to get the equation, that is the polynomial from the roots?

Answer 5

anyhow... if we had a quadratic whose roots are x = −1, x = 2, and x = 4.

that means \(\bf x = −1\implies {\color{blue}{ x+1}}=0\quad x = 2\implies {\color{blue}{ x-2}}=0\quad x = 4\implies {\color{blue}{ x-4}}=0\\ \quad \\

(x+1)(x-2)(x-4)=\textit{the original quadratic equation}\)

Answer 6

if that quadratic equation has those roots, that means is DEFINED at those points

Answer 7

now, for a rational expression, a vertical asymptote occurs, at the ZEROS OF THE DENOMINATOR

so, if we use that same quadratic expression, and use it as our denominator in a rational expression, those same ZEROS or roots of that expression
will become the vertical asymptotes of the rational expression

an asymptote is where the rational function is UNDEFINED
because it denominator is 0, and a fraction with 0 in the denominator is UNDEFINED

Answer 8

Aw thank you! I knew that they could be defined but i didnt know where to start to prove that. But thats also where im confused. I understand that they can be defined by those roots but it says to create a scenario where edward is correct and he says that they are NOT defined by those roots. That is where i get confused

Answer 9

\(\bf x = -1\implies {\color{blue}{ x+1}}=0\quad x = 2\implies {\color{blue}{ x-2}}=0\quad x = 4\implies {\color{blue}{ x-4}}=0\\ \quad \\

(x+1)(x-2)(x-4)=\textit{the original quadratic equation}\\ \quad \\
\textit{DEFINED at }x=-1\quad x=2\quad x=4\\ \quad \\
----------------------------\\
\cfrac{x}{\textit{the original quadratic equation}}\quad or\quad \cfrac{7}{\textit{the original quadratic equation}}\\ \quad \\
\textit{are UNDEFINED at }x=-1\quad x=2\quad x=4\)

Answer 10

well, it has 3 roots...so is really a "cubic" not a quadratic in this case =)

Answer 11

\(\bf x = -1\implies {\color{blue}{ x+1}}=0\quad x = 2\implies {\color{blue}{ x-2}}=0\quad x = 4\implies {\color{blue}{ x-4}}=0\\ \quad \\

(x+1)(x-2)(x-4)=\textit{the original cubic equation}\\ \quad \\
\textit{DEFINED at }x=-1\quad x=2\quad x=4\\ \quad \\
----------------------------\\
\cfrac{x}{\textit{the original cubic equation}}\quad or\quad \cfrac{7}{\textit{the original cubic equation}}\\ \quad \\
\textit{are UNDEFINED at }x=-1\quad x=2\quad x=4\)

Answer 12

You my friend are a life saver! Thank you so much. I understand now (:

Answer 13

yw