Video: Finding the Second Derivative of a Function Defined by Parametric Equations

Find the horizontal and vertical asymptotes of 𝑓(π‘₯) = 1/(π‘₯ βˆ’ 1) βˆ’ 2.

02:22

Video Transcript

Find the horizontal and vertical asymptotes of 𝑓 of π‘₯ equals one over π‘₯ minus one minus two.

We begin by recalling the definition of a horizontal and vertical asymptote. We say that the line 𝑦 equals 𝐿 is a horizontal asymptote of a function 𝑦 equals 𝑓 of π‘₯ if the limit as π‘₯ approaches either negative or positive ∞ of 𝑓 of π‘₯ is equal to 𝐿. Similarly, π‘₯ equals π‘Ž is a vertical asymptote of a function 𝑦 equals 𝑓 of π‘₯ if any of the six statements are true. That is, the limit as π‘₯ approaches π‘Ž from the left of the function is either positive or negative ∞. The limit as π‘₯ approaches π‘Ž from the right of the function is either positive or negative ∞. Or the limit as π‘₯ approaches π‘Ž of the function is positive or negative ∞.

Now, our function 𝑓 of π‘₯ is equal to one over π‘₯ minus one minus two. So we’ll begin by evaluating the limit as π‘₯ approaches ∞ of one over π‘₯ minus one minus two. And in fact, very little manipulation is required here. As π‘₯ gets sufficiently large, one over π‘₯ minus one approaches zero. And so the limit as π‘₯ approaches ∞ of one over π‘₯ minus one minus two is zero minus two, which is simply negative two. And in fact, as π‘₯ approaches negative ∞, one over π‘₯ minus one still approaches zero. So the limit as π‘₯ approaches negative ∞ of 𝑓 of π‘₯ is equal to negative two. And we find the equation 𝑦 equals negative two is a horizontal asymptote of our function.

Next, we’ll consider the vertical asymptotes. Less formally, we say that, with rational functions, that’s those written as the quotient of two functions, vertical asymptotes are the lines which correspond to the zeros of the denominator of the function. So in fact, we’re going to begin by subtracting two from one over π‘₯ minus one by creating a common denominator. We multiply two over one by π‘₯ minus one over π‘₯ minus one. And then, we can subtract the numerator. So we have one minus two times π‘₯ minus one over π‘₯ minus one. We simplify. And we find that 𝑓 of π‘₯ can be written as three minus two π‘₯ over π‘₯ minus one.

We’re interested in the values of π‘₯ so that our denominator π‘₯ minus one is equal to zero. Well, if we solve this equation for π‘₯ by adding one to both sides, we find that π‘₯ is equal to one. And so we found the horizontal and vertical asymptotes of 𝑓 of π‘₯ equals one over π‘₯ minus one minus two. The horizontal asymptote is 𝑦 equals negative two. And the vertical asymptote is π‘₯ equals one.

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