find an asymptote to the curve y^2(a + 2x) = x^2(3a - x)

Question

the answer given is x = - a /2 . Please explain how? 

Answer 1

y²(a+2x) = x²(3a-x)

y² = x²(3a-x)/(a+2x)

y = x√[(3a-x)/(a+2x)]

The Domain of a function is the interval of x-values that the graph uses to get all of its points on there. So basically it tells us all of the x-values that we can plug into the function, and can actually solve for y to get a real number. 

A lot of times, the Domain of a function will be "all real numbers," because there aren't too many things that interfere with the Domain. The two I'm sure you've seen already is that

1. You can't take the square root of negative numbers

• If you see a √(-1) when solving for y, then that (x, y) won't show up on the graph

2. You can't divide by 0

• If the x-value you plug in results in dividing by 0 to solve for y, that (x, y) won't show up

There are a few more, but between these two that apply to this function, only the second one shows up as a vertical asymptote. So even though we're not allowed to divide by 0 for the function...... to find out where that asymptote is, we have to ask ourselves where the denominator is 0. So in other words, set the denominator equal to 0 and solve for x.

a + 2x = 0

2x = -a

x = -a/2

Hope this helps,

Frank

Answer 2

y²(a + 2x) = x²(3a - x)

 

y² = x²(3a-x)/(a+2x)

 

Can't divide by zero, so we can't have:

 

a + 2x = 0

 

Solve for x.