Consider the function 𝑦 equals three 𝑥 divided by five 𝑥 plus seven. By considering the point at which the denominator equals zero, find the domain of the
function. To find the range of the function, a handy trick is to divide the numerator and
denominator of three 𝑥 over five 𝑥 plus seven through by 𝑥. What expression does this give us? Now taking the limit of this expression as 𝑥 tends to ∞ will give us the value of
𝑦, which is not the range of the original function. Use this to state the range of the function. Hence, state the equations of the two asymptotes.
So here is our equation, and it first asks us to consider the point at which the
denominator equals zero to find the domain. So our denominator is five 𝑥 plus seven. So we need to set this equal to zero. So to solve for 𝑥, we must first subtract seven from both sides of the equation. And we have that five 𝑥 equals negative seven.
So now our last step to solve for 𝑥 would be to divide both sides of the equation by
five, leaving us with 𝑥 equals negative seven-fifths. So why are we supposed to set the denominator equal to zero?
Well, if we have a denominator of zero, then we have something that’s undefined; so
it doesn’t work. So this 𝑥 equals negative seven-fifths means it cannot happen. That is a value we cannot put on the denominator because it will actually give us
zero on the denominator. And that can’t happen; we don’t want it to. Our function would be undefined there. So this means we can have any value of 𝑥 for the domain except for negative
seven-fifths. And the way to write this would be 𝑥 is in the all real numbers comma 𝑥 can’t be
equal to negative seven-fifths. So it’s all values of 𝑥 in a real number system besides negative seven-fifths.
Next, to find the range, they give us a handy trick. It says to divide the numerator and denominator by 𝑥. So here is our equation. Three 𝑥 is the numerator, and five 𝑥 plus seven is the denominator. So we are supposed to divide each of these by 𝑥. So when we take the numerator and divide it by 𝑥, the 𝑥s cancel. And we’re left with three.
Now, for the next one, we’re gonna have to separate these. We can’t just take out an 𝑥 out of a binomial. The five 𝑥 plus seven is a binomial. So instead, we’ll have to separate these and divide each one by 𝑥. So five 𝑥 over 𝑥, the 𝑥s cancel, and we have five. But seven over 𝑥 does not simplify, so we leave it this way.
So putting the numerator back where the numerator goes and the denominator where the
denominator goes, that will give us the expression. So the expression we have is three divided by five plus seven over 𝑥.
Next, it says now taking the limit of this expression as 𝑥 tends to ∞ will give us
the value of 𝑦 at which is not the range of the original function. So here is the expression we had, and it says to take the limit of the expression as
𝑥 tends to ∞. So 𝑥 was going towards ∞. We can just put the ∞ there. So think about it. We really can’t divide by ∞; that’s why it’s called a limit.
So seven divided by ∞, ∞ is a very very very very large number. And if we take a number like seven and divide it by a very large number, we’re gonna
get a very very tiny fraction. So the bigger the number gets, the smaller the fraction we get. So if we get smaller and smaller and smaller, what number are we getting closer
to? Zero. So we can replace the seven over ∞ with zero.
But five plus zero on the denominator is just five, so we’re left with
three-fifths. So it taught us that this would not be in the range. So this means we can have all values of 𝑦 except for this three-fifths. And the proper way to write this would be 𝑦 is in the real numbers, but 𝑦 cannot be
equal to three-fifths.
Lastly, it says, hence, state the equations of the two asymptotes. First, what’s an asymptote? It’s a point on the graph that the function approaches but never touches. So what were the values that we found we actually couldn’t land on? They were in the domain, and they were in a range. That was the negative seven-fifths for 𝑥 and the three-fifths for 𝑦. These will be the asymptotes of the lines that the function will approach but not
touch, leaving us with the equations 𝑦 equals three-fifths and 𝑥 equals negative