Video: Sketching Graphs of Rational Functions

Which of the following is the equation of the rational function graphed below? [A] 𝑓(π‘₯) = (π‘₯ + 4)(π‘₯ + 1)(π‘₯ βˆ’ 1)(π‘₯ βˆ’ 3) [B] 𝑓(π‘₯) = (1/14)(π‘₯ + 4)(π‘₯ + 1)(π‘₯ βˆ’ 1)(π‘₯ βˆ’ 3) + 0.5 [C] 𝑓(π‘₯) = (π‘₯ + 4)(π‘₯ + 1)(π‘₯ βˆ’ 1) + 0.5 [D] 𝑓(π‘₯) = (π‘₯ βˆ’ 4)(π‘₯ + 4)(π‘₯ + 1)(π‘₯ βˆ’ 1)(π‘₯ βˆ’ 3) + 0.5

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Video Transcript

Which of the following is the equation of the rational function graph below? Is it (A) 𝑓 of π‘₯ is π‘₯ plus four times π‘₯ plus one times π‘₯ minus one times π‘₯ minus three? (B) 𝑓 of π‘₯ is one fourteenth π‘₯ plus four times π‘₯ plus one times π‘₯ minus one times π‘₯ minus three plus 0.5. (C) 𝑓 of π‘₯ is π‘₯ plus four times π‘₯ plus one times π‘₯ minus one plus 0.5. Or (D) 𝑓 of π‘₯ is π‘₯ minus four times π‘₯ plus four times π‘₯ plus one times π‘₯ minus one times π‘₯ minus three plus 0.5.

We’ve been given the graph of a rational function 𝑓 of π‘₯. Before looking at our options for the equation of this graph, let’s just think what we do know about it. It’s a sort of W shape. It has one, two, three turning points. Now, one thing we know about polynomial functions is that the maximum number of turning points of a polynomial function is always one less than the degree of the function. And this means the degree of our function is at least four. It’s four and above.

Now, the degree of the function also tells us the general shape. We’ve already said that the degree is four or above. Now, we know that a function with a degree of four has either a W or an M shape. Now, the sign of the coefficient of the leading power of π‘₯ will tell us if it’s a W or an M. And the shape of a graph with a degree of five is as shown. Once again, this very much depends on the sign of the highest power of π‘₯ . Now, we said our graph has a W shape, so its degree must indeed be equal to four.

We’re now going to go back to the equations that we’ve been given in this question and see if we can eliminate any. Let’s look at (A). 𝑓 of π‘₯ is π‘₯ plus four times π‘₯ plus one times π‘₯ minus one times π‘₯ minus three. Now, let’s imagine we were to distribute these parentheses or expand our brackets. Eventually, we would end up multiplying π‘₯ by π‘₯ by π‘₯ by π‘₯, giving us π‘₯ to the fourth power. So the function could indeed be function (A).

Similarly, distributing the parentheses for function (B) also gives us π‘₯ to the fourth power as the highest power of π‘₯. So it could be function (B). If we were to distribute the parentheses for function (C), however, we’d be multiplying eventually π‘₯ by π‘₯ by π‘₯. Now, that actually gives us π‘₯ cubed. And so we can eliminate function (C) from our question.

Finally, if we were to distribute the parentheses from function (D), we’d be multiplying π‘₯ by π‘₯ by π‘₯ by π‘₯ and then by π‘₯ again, giving us π‘₯ to the fifth power. And so we’re going to eliminate function (D).

Now, we’re simply left with choosing between function (A) and function (B). And so next, we could consider the roots or the zeros of our equation. The roots of our equation correspond to the π‘₯-intercepts of our graph. Now, actually, these aren’t particularly nice to define from the graph. If we try to read the π‘₯-intercepts off of our graph, we get a roughly π‘₯ is equal to negative 3.9, negative 1.3, 1.3, and roughly 2.8. Let’s compare these to our final two equations.

To find the roots or the zeros of our equation, we set the equation equal to zero and solve for π‘₯. So for equation (A), that’s π‘₯ plus four times π‘₯ plus one times π‘₯ minus one times π‘₯ minus three equals zero. Now, each binomial, each expression inside our parentheses, is itself a number. And when we multiply these four numbers together, we get zero. So what does that tell us about any one of these numbers? Well, it tells us that at least one of these numbers must itself be equal to zero. So we could say that either π‘₯ plus four is equal to zero, π‘₯ plus one is equal to zero, π‘₯ minus one is equal to zero, or π‘₯ minus three is equal to zero.

Solving for π‘₯ in our first equation, we subtract four, in our second, we subtract one, and so on. And we find the zeros of this equation to be π‘₯ equals negative four, π‘₯ equals negative one, π‘₯ equals one, and π‘₯ equals three. Now, we actually see that these do not correspond to the π‘₯-intercepts that we’ve read from our graph. And so (A) cannot be the correct graph. And it therefore must be (B).

Now, we would have a little more of a job trying to solve the equation of fourteenth π‘₯ plus four times π‘₯ plus one times π‘₯ minus one times π‘₯ minus three plus 0.5 equals zero. We’d have to distribute our parentheses, refactor, and then solve for π‘₯. But there is one other hint that can tell us that graph (B) is indeed the correct graph. We know that, for a polynomial function, the constant value will tell us the value of the 𝑦-intercept. So let’s imagine we were distributing our parentheses. What would our constant be?

Well, if we were just multiplying the constant inside our parentheses, we get 12. We then see that we multiply that by a fourteenth and add 0.5. That gives us 1.357 and so on. And whilst it’s difficult to read from the graph, we can see that this is indeed the value of our 𝑦-intercept. And so the equation of the rational function we’ve been given is (B). It’s 𝑓 of π‘₯ equals a fourteenth times π‘₯ plus four π‘₯ plus one times π‘₯ minus one times π‘₯ minus three plus 0.5.

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