Question

Algebra 2 help please. I will give metal! I am kinda confused with this one.

Answer 1

Identify the vertical asymptotes of f(x) = 2 over quantity x squared plus 3 x minus 10

Answer 2

ChristopherToni I thought you were going to sleep that is why I didn't ask you one the other one. I felt bad keeping you awake.

Answer 3

No worries; it usually takes me a few minutes to really force myself to go to bed. I'll help you with this last one before really departing for the night.

Answer 4

Ok Thank you!!!!!!

Answer 5

The thing you need to be careful about with vertical asymptotes is this: the general criterion for checking if one is a VA or not is finding the values where the denominator is equal to zero; i.e. the values where you're dividing by zero (because dividing by zero is bad). However, there may be terms in the numerator as well that may be zero when the denominator is zero). For the most part, this phenomenon generates what is called a hole in the graph. For instance, at first glance for the function \[\large f(x) = \frac{x^2-3x+2}{x^2-5x+6} = \frac{(x-2)(x-1)}{(x-3)(x-2)}\]
one may think that \(x=3\) and \(x=2\) are the vertical asymptotes, but then you would note that the numerator also has the factor of \(x-2\); hence when you cancel this out, this creates a hole in the graph of this function and thus \(x=2\) is no longer a candidate for being a vertical asymptote. Therefore, in this example, \(x=3\) would be the only vertical asymptote.

Now, to answer your question, note that
\[\large \frac{2}{x^2+3x-10} = \frac{2}{(x+5)(x-2)}\]
Can you see what the vertical asymptotes are now? :-)

I hope the explanation I provided above made sense! :)

Answer 6

I see what you are saying but that is not an answer chouce

Answer 7

x = 5 and x = 2

x = 5 and x = –2

x = –5 and x = 2

x = –5 and x = –2

Answer 8

those are my answer choices but I am about to get kicked off the computer. Maybe you can help me later? If so please message me when the best time is for you to come help. Sorry
Thanks again though.

Answer 9

Again, to find the vertical asymptotes, you want to see where the denominator of your function is zero. In this case, we don't have to worry about there being any holes since the numerator of your fraction is just a constant. So, when is \(\large (x+5)(x-2)=0\)? The solutions you get are in fact the equations of your vertical asymptotes. Does this make sense? :-)

Answer 10

so they are -5 and +2?

Answer 11

Yes, they are. :-)

Answer 12

Yay! Thanks again!!! Now go to sleep haha. Bye!

Answer 13

I'll be back online around 1:30PM PST in case you have any more questions you need help with. :-)

Answer 14

ok thank you