Question Video: Sketching Curves from their Roots

Which of the following is the graph of 𝑓(π‘₯) = (π‘₯ + 1)Β²(π‘₯ βˆ’ 2)(π‘₯ βˆ’ 1)? [A] Graph A [B] Graph B [C] Graph C [D] Graph D [E] Graph E


Video Transcript

Which of the following is the graph of 𝑓 of π‘₯ is equal to π‘₯ plus one all squared times π‘₯ minus two multiplied by π‘₯ minus one? Option (A), (B), (C), (D), or (E).

We’re given five possible sketches of the curve 𝑦 is equal to 𝑓 of π‘₯. And we could see if we were to multiply our expression for 𝑓 of π‘₯, we would get a polynomial. So to answer this question, we’ll use what we know about sketching polynomials. The first thing we can do when we’re sketching any function is substitute π‘₯ is equal to zero to find the 𝑦-intercept. So we’ll substitute π‘₯ is equal to zero into our function 𝑓 of π‘₯. So we substitute π‘₯ is equal to zero into 𝑓 of π‘₯. We get zero plus one all squared times zero minus two multiplied by zero minus one. And if we evaluate this expression, we get two.

Now, two is a positive number. So our 𝑦-intercept must be above the π‘₯-axis. And if we look in option (E), we can see that the 𝑦-intercept is below the π‘₯-axis. It can’t be option (E). The second thing we should look for is the end behavior of our curve. For polynomial functions, we can determine the end behavior of a curve by looking at the leading term. In our case, we’re not given an expanded form for our graph. So we don’t know the leading term. We would need to multiply this out.

However, we only need the highest power of π‘₯. If we were to do this, we would see we would just get π‘₯ to the fourth power. So we can determine the end behavior of our curve 𝑦 is equal to 𝑓 of π‘₯ just by looking at the end behavior of π‘₯ to the fourth power. We know for large positive values of π‘₯, π‘₯ to the fourth power is positive. And for very negative values of π‘₯, it’s also positive.

Also, remember, we showed that the 𝑦-intercept must be at two. So we can clear this to avoid confusion and add this to our sketch. But now we can see something interesting. Options (B), (C), and (D) don’t have the same end behavior as our curve. So none of these can be the correct sketch. Therefore, by elimination, option (A) must be the correct sketch. It’s also worth pointing out we could’ve sketched this directly from the function 𝑓 of π‘₯.

The third thing we want to do when sketching a polynomial curve is find the π‘₯-intercepts. π‘₯-intercepts will be where our function’s output is equal to zero. And if we factor our polynomial, the products of factors being equal to zero means that one of our factors must be equal to zero. So we just solve each factor equal to zero. And this gives us our three π‘₯-intercepts: negative one, two, and one. And we can add these to our sketch.

The last thing we need to know is the shape of our function near the intercepts. And we recall we can do this by looking at the multiplicity of the factor. For example, when π‘₯ is equal to two and when π‘₯ is equal to one, the multiplicity of these two factors is one. We could find a direct approximation by substitution. However, this is not necessary. We know this is going to behave like a linear function, a straight line.

Look at π‘₯ is equal to two however. We know the end behavior of our curve near this point. So this slope at π‘₯ is equal to two must be positive. It must be a positive linear function. But then the opposite must be true at one, because our curve can’t intersect the π‘₯-axis anywhere else. It must be a negative slope linear function. And this means we can fill in some of our curve. It will look something like this.

When π‘₯ is equal to negative one however, when we look at our factor, we can see its multiplicity is two. This means near the value of negative one, our curve will look like some multiple of π‘₯ plus one all squared. Now, we could find the value of 𝐴 by substituting negative one into the remaining part of our function. However, it’s not necessary. We know that 𝐴 must be positive just by looking at our curve. This is the only way that the shape of our curve could make sense. So 𝐴 is positive. And we can fill in the rest of our curve. Our curve is just going to touch the π‘₯-axis at the point negative one. And this confirms that option (A) was the correct choice.

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