Solving f(x)=0: Horizontal Asymptote

[Boom101]

Homework Statement

f(x) = 2 / x + 3

Homework Equations

None

The Attempt at a Solution

Nvm I'm an idiot. Y=0 is a horizontal asymptote

 

[symbolipoint]

Going by the assumption that you mean f(x) = \frac{2}{x} + 3 , the domain excludes x=0. The graph must show an asymptote for the line x=0. As x approaches 0 from the left, f(x) tends to -\infty; as x approaches 0 from the right, f(x) approaches +\infty. Consider carefully what happens to the function as x itself decreases without bound, and what happens as x increases without bound!

 

[Mark44]

Or the OP might have meant f(x) = 2/(x + 3), but isn't aware of the need for parentheses to distinguish this function from y = (2/x) + 3.

Both functions are asymptotic to the x-axis, but have different vertical asymptotes.

 

[Boom101]

Is it absolutely necessary to find out the behavior of x -> a if you calculate whether the interval increases/decreases and the concavity?

 

[Boom101]

And i did mean 2 / (x+3). I was just copying what was down on the page, but I know my original post was wrong. I was looking for the y intercept, which there isn't, correct? I'm just wondering if on top of finding if the interval increase/decrease and concavity I should also find the behavior of the limit. Thanks

 

[Mark44]

Is it absolutely necessary to find out the behavior of x -> a if you calculate whether the interval increases/decreases and the concavity?

Let's get the terminology right so that people can understand what you're trying to say. The interval doesn't decrease or increase, but there are intervals on which the function increases or decreases. And there are intervals on which the graph of the function is concave up or concave down.

Knowing where the function is increasing or decreasing and where its graph is concave up or concave down will give you an idea of what the limits near the vertical asymptotes, so you could probably get away with not doing that work. OTOH, doing it gives you a check against your other work, so if you get different results, that means you did something wrong.

And i did mean 2 / (x+3). I was just copying what was down on the page, but I know my original post was wrong. I was looking for the y intercept, which there isn't, correct? I'm just wondering if on top of finding if the interval increase/decrease and concavity I should also find the behavior of the limit. Thanks

Right, there is no y-intercept. Finding the behavior around the vertical asymptote and as x gets large or very negative can be done quickly. When x is near -3, the denominator is close to 0, but the values will all be positive on one side of -3 and will all be negative for values on the other side of -3.

When x is very large, the expression 2/(x + 3) will be close to 0 (but positive). When x is very negative, this expression will also be close to 0, but will be negative. These say something about how the graph of the function approaches zero for very large or very negative values of x.