Question Video: Identifying Graphs of Higher-Degree Polynomials | Nagwa Question Video: Identifying Graphs of Higher-Degree Polynomials | Nagwa

Question Video: Identifying Graphs of Higher-Degree Polynomials

Consider the following graphs. Which of these is the graph 𝑓(𝑥) = (𝑥 + 1)²/((𝑥 − 1)(𝑥 + 3))? [A] Graph (a) [B] Graph (b) [C] Graph (c) [D] Graph (d).

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Video Transcript

Consider the following graphs. Which of these is the graph 𝑓 of 𝑥 equals 𝑥 plus one squared divided by 𝑥 minus one times 𝑥 plus three?

So here we have a rational function. It’s a fraction, and rational functions have asymptotes. The denominators have to be nonzero. Because if you take anything and divide by zero, it will be undefined, so we don’t want that! So what would make our denominator zero would be one and negative three. So how did we get that? We should take the two factors that are on the denominator and set them equal to zero.

So now we need to add one to our left equation and subtract three from our right equation. So if we would plug in one for 𝑥, just looking at the denominator, we would get zero times negative two and zero times anything is zero. So if we plugged in one, that left expression would be zero, and if we plug in negative three, that right expression will be zero. But since you multiply them together, either one of those would give us a final answer of zero on our denominator, and again we don’t want that.

So here we can see every single graph actually has the asymptotes at negative three and one. So that doesn’t eliminate any of our options. So our next step would be to see what the graphs are doing when they’re really close to these asymptotes, and then we’ll know, so let’s think about it. If we have two asymptotes at negative three and one, that means we will never actually be able to land on those values. It doesn’t work for our function. So we can plug in numbers really close to them and see what happens.

So we can plug in values to the left of negative three, so just a little bit more negative than negative three, so negative 3.01, negative 3.0009, and so on. And then we are also wanna plug in numbers to the right of that asymptote, so more positive, so negative 2.999, negative 2.9999999, and so on. And then the same thing for one, numbers are really close to one that are a little bit less than that and a little bit greater than that.

So let’s first begin with numbers to the left of our furthest left asymptote, 𝑥 equals negative three. So for example negative 3.01, let’s plug that in. So taking our function instead of 𝑥, we’re replacing it with negative 3.01. And we get 100.75, so something pretty high up. So now let’s plug in a number that’s even closer to 𝑥 equals negative three. So instead of negative 3.01, how about 𝑥 equals negative 3.001? And when we plug that in, we get 1000.75, so way higher than that!

So it must be going up in the 𝑦-direction, becoming much greater. So we can go ahead and exclude option a because just to the left of our asymptote of 𝑥 equals negative three, it should be going up, increasing. So now let’s look to the right of that asymptote, so we can go ahead and exclude option a because it should be increasing; the number should be coming pretty pretty large for 𝑦 as we get really close to that asymptote of 𝑥 equals negative three.

So now let’s plug in numbers to the right of it such as 𝑥 equals negative 2.9 and 𝑥 equals negative 2.99. So at 𝑥 equals negative 2.9, we get a negative number, negative 9.26. So now let’s plug in the number that’s even closer to that dash line, negative 2.99, and we get a number even more negative, negative 99.25. So this is decreasing as we approach 𝑥 equals negative three from the right side of it.

And all three of our remaining options do that; however, a few of them are a little bit higher than the others, so why don’t we plug in negative one for 𝑥 and see where we should land? So plugging in negative one, we get zero squared divided by negative two times two. And zero divided by negative four is zero. So at 𝑥 equals negative one, we should be at zero for 𝑦, which would mean our answer would have to be graph c.

However just to double check, let’s keep checking the values around 𝑥 equals one for the asymptote and make sure. So the numbers on the left side of the asymptote 𝑥 equals one or a little less than one, so let’s plug in 𝑥 equals 0.9 and 0.99. When we plug in 0.9, we get negative 9.26. And now we’re plugging in a number a little bit closer to that asymptote 𝑥 equals one, and we get negative 99.25. So we are decreasing again, which is what’s happening in our graph.

So that’s good! So finally we will plug in numbers to the right of that asymptote 𝑥 equals one, so little bit bigger than one. So we will plug in 1.01 and 1.001. When we plug in 1.01, we get 100.75, so a pretty big number. And now a number that’s closer to that line, 1.001, we get an even larger number, so it will be increasing, which is exactly what’s happening in this option of graph c. So this graph is what represents our function.

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