Video: Sketching Graphs of Rational Functions

Which of the following is the equation of the rational function graphed below? [A] 𝑓(π‘₯) = π‘₯Β² + 3π‘₯ βˆ’ 8 [B] 𝑓(π‘₯) = π‘₯⁴ + 3π‘₯Β³ βˆ’ 2π‘₯Β² βˆ’ 8π‘₯ + 1 [C] 𝑓(π‘₯) = π‘₯Β³ + 3π‘₯Β² βˆ’ 2π‘₯ βˆ’ 8 [D] 𝑓(π‘₯) = π‘₯Β³ βˆ’ 2π‘₯

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Video Transcript

Which of the following is the equation of the rational function graphed below? Is it (A) 𝑓 of π‘₯ is π‘₯ squared plus three π‘₯ minus eight? (B) 𝑓 of π‘₯ is π‘₯ to the fourth power plus three π‘₯ cubed minus two π‘₯ squared minus eight π‘₯ plus one. Is it (C) 𝑓 of π‘₯ is π‘₯ cubed plus three π‘₯ squared minus two π‘₯ minus eight? Or (D) 𝑓 of π‘₯ is π‘₯ cubed minus two π‘₯.

Let’s begin by looking at the graph we’ve been given. We’re told it’s the graph of a rational function defined by 𝑓 of π‘₯. And the options for 𝑓 of π‘₯ that we’ve been given are all polynomials. So let’s think about what we know about the shape of our graph to start with. The graph itself has one, two turning points. And we know that a polynomial of degree 𝑛 can have up to 𝑛 minus one turning points. And so we see that the degree of our function must be three or above.

Now, this means we can instantly eliminate graph A. The degree of the function (A) is two. It’s a quadratic function. And in fact we know that the graph of quadratic functions look a little bit like these. They’re parabolas, and so we’re left with three functions. The degree of one of our functions is four and the degree of the other two is three. So next, we’ll consider the shape of the graphs. We know that the shape of the graph of a function with a degree of three is as shown, whereas the shape of the graph of a function with degree four is a W or an M shape. Now the shape that we end up with depends on the sign of the coefficient of the highest power of π‘₯.

If we compare the graph of our function to the general shape, we see that it must have a degree of three. And so we can eliminate graph (B), which has a degree of four. And so now, we’re left with two functions. (C) is 𝑓 of π‘₯ is π‘₯ cubed plus three π‘₯ squared minus two π‘₯ minus eight. And (D) is 𝑓 of π‘₯ equals π‘₯ cubed minus two π‘₯. Well, there are two things we’re going to consider. And those are the π‘₯- and 𝑦-intercepts of our graph.

Firstly, we know that, with a polynomial function, the value of the constant tells us the value of the 𝑦-intercept. Well, in function (C), its constant is negative eight whereas in function (D), its constant is zero. And so the graph of function (C) must pass through the 𝑦-axis at the point with coordinates zero, negative eight, whereas the graph of function D must pass through the point zero, zero. That’s the origin.

Now, if we look at the graph of our function, we see it does indeed pass through the point zero, negative eight. And so we eliminate function (D) from our problem. But how could we have checked using the π‘₯-intercepts? Well, to find the value of the π‘₯-intercepts, we set the equations equal to zero and then solve for π‘₯. Equation (C) is quite tricky to solve. So let’s have a look at equation (D). We solve by factoring the expression on the left-hand side. And when we do, we get π‘₯ times π‘₯ squared minus two.

Now, for this statement, π‘₯ times π‘₯ squared minus two equals zero, to be true, either π‘₯ itself must be equal to zero or π‘₯ squared minus two must be equal to zero. And if we solve this second equation for π‘₯, we find π‘₯ is equal to plus or minus root two. And so the π‘₯-intercepts of the function π‘₯ cubed minus two π‘₯ are zero and plus or minus root two. Once again, we see that these do not correspond to the zeros or the π‘₯-intercepts of the function given.

And so the equation of the rational function graphed is (C). It’s 𝑓 of π‘₯ equals π‘₯ cubed plus three π‘₯ squared minus two π‘₯ minus eight.

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