Continuity Regarding a Hole and Vertical Asymptote

[Jason76]

For this type of function there would be a hole:

Ex 1.

$\displaystyle y = \dfrac{x^{2} - 5}{x - 5}$ which reduces to

$\displaystyle y = \dfrac{(x + 5)(x - 5)}{x - 5}$

would have a hole at $\displaystyle x = 5$

to find the missing y value, then you would cancel out the "terms that cancel out" and solve for y.

When $\displaystyle x = 5$

$\displaystyle y = 5^{2} + 5 = 30$

However this next function would have a vertical asymptote at $\displaystyle x = 5$:

Ex 2.

$\displaystyle \dfrac{(7x - 5)}{x - 5}$

Because it has 0 in the bottom when x = 5, and there is nothing to cancel out. Is this all right?

 

[lookagain]

For this type of function there would be a hole:

Ex 1.

$\displaystyle y = \dfrac{x^{2} - 25}{x - 5} \ \ \$ if the numerator is supposed to be $\displaystyle \ \ "x^2 - 25."$

$\displaystyle y = \dfrac{(x + 5)(x - 5)}{x - 5}$

would have a hole at $\displaystyle x = 5$

to find the missing y value, then you would cancel out the "terms that cancel out" and solve for y.

When $\displaystyle x = 5$

$\displaystyle y = 5^{2} + 5 = 30$ . . . . . . No. When $\displaystyle x \ne 0, \$ the (x - 5) factors reduce to ones, and you end up with y = x + 5. Therefore, when x = 5, there is a hole at the point (5, 10).

However this next function would have a vertical asymptote at $\displaystyle x = 5$:

Ex 2.

$\displaystyle \dfrac{(7x - 5)}{x - 5}$

Because it has 0 in the bottom when x = 5, and there is nothing to cancel out. Is this all right? . . . No, it is not right for example 1.

.

 

[Bob Brown MSEE]

[h=2]Neither has Hole, both have Vertical Asymptote[/h]unless the 5 in numerator is actually 25

 

[Jason76]

.

Ok, I see what your saying.

Neither has Hole, both have Vertical Asymptote

What do you mean?

 

[Bob Brown MSEE]

"Neither has Hole, both have Vertical Asymptote"

 

[lookagain]

For this type of function there would be a hole:

Ex 1.

$\displaystyle y = \dfrac{x^{2} - 5}{x - 5}$ <-----------

Jason76, please confirm if the function is actually supposed to be $\displaystyle \ y = \dfrac{x^2 - 25}{x - 5}. \ \ \$ If so, then stick with my points in post # 2.

 

[Jason76]

Jason76, please confirm if the function is actually supposed to be $\displaystyle \ y = \dfrac{x^2 - 25}{x - 5}. \ \ \$ If so, then stick with my points in post # 2.

$\displaystyle \ y = \dfrac{x^2 - 25}{x - 5}. \ \ \$ is the function.

Neither has Hole, both have Vertical Asymptote

unless the 5 in numerator is actually 25

I understand your meaning now.