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Graphs of polynomials: Challenge problems

Solve challenging problems that tackle the relationship between the features of a polynomial and its graph.

What you should be familiar with before taking this lesson

What you will do in this lesson

Now that we have learned about the features of the graphs of polynomial functions, let's put that knowledge to use!
In this set of problems, the equations of the polynomials are not completely given. This way, they force us to focus on a specific feature of the polynomial's graph.
Good luck!
1*) Which of the following could be the graph of y, equals, a, x, cubed, plus, b, x, squared, plus, c, x, plus, 2, where a, b, and c are real numbers?
Choose 1 answer:
Choose 1 answer:

2*) Which of the following could be the graph of y, equals, minus, 2, x, start superscript, 5, end superscript, plus, p, left parenthesis, x, right parenthesis, where p, left parenthesis, x, right parenthesis is a fourth degree polynomial?
Choose 1 answer:
Choose 1 answer:

3*) Which of the following could be the graph of y, equals, k, left parenthesis, x, minus, 2, right parenthesis, start superscript, m, end superscript, left parenthesis, x, plus, 1, right parenthesis, start superscript, n, end superscript, where k is a real number, m is an even integer, and n is an odd integer?
Choose all answers that apply:
Choose all answers that apply:

Want to join the conversation?

  • mr pink green style avatar for user Ahmed Bahgat
    At the last question, I think all of the graphs are wrong. if we look at the leading term, it will be the the product of an even number and an odd number, which means that the leading term will always be even (odd * even = even). Therefore, both of the ends of the graph will be at the same side of the x-axis.
    (5 votes)
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    • blobby green style avatar for user Vanessa Eluma
      The first term has an even exponent and the next term has an odd one, thus you would add the exponents NOT multiply them. As such, the degree of this polynomial will be an odd integer, or "odd + even = odd".

      For instance, imagine if m = 2 and n = 3, and you expanded the binomials. Then, you would end up with a polynomial with a degree of five.
      (3 votes)
  • female robot grace style avatar for user Sarah Fowler
    In the first question, the y intercept is 2, but the graph of C is of a negative odd-degree polynomial, isn't it? Why is that the answer?
    (6 votes)
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    • blobby green style avatar for user J T
      Yes it is a neg. odd-degree poly. The question gives very little info, the only thing we know is the y intercept is 2. So that is all we can go off. Since graph C is the ONLY graph with the line crossing the y intercept at 2, that has to be our answer since we have nothing else to go off.
      Now back to the neg. odd-degree poly., If a < 0 (e.g. -7) does the y intercept change? No, and the graph fits, so from the graph we can conclude that 'a' is a negative number.
      Hope that clears it up for you! Good Luck
      (2 votes)
  • blobby green style avatar for user lori keller
    Where do I look on the site to find information about how to solve this question?

    Sketch a quartic function with a leading coefficient of -2, with two negative zeros and two complex roots.
    (3 votes)
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    • aqualine ultimate style avatar for user Alec Traaseth
      I would just create a polynomial that meets those requirements, starting with the factored form. Quartic means it will have 4 linear factors. The easiest complex roots to deal with are (x+i) and (x-i), and you need to have the conjugate for whichever complex root you pick, so those two satisfy that requirement. Then you just pick two negative zeroes, let's use -1 and -2. Lastly, you can multiply on the outside by -2 to ensure your leading coefficient is -2.

      From that, we get P(x) = -2(x+1)(x+2)(x+i)(x-i). To sketch it, you need to have the correct end behavior and all intercepts. The x intercepts will be at -1 and -2, since we chose those for our zeroes. Since our function is of even degree, it will go in the same direction on both sides of the graph, and the negative out front means it will be negative in both the positive infinity direction and the negatie infinity direction. Our y intercept comes from the constant term after multiplying it out, which will be 1 x 2 x -2 = -4. Just hit all those points, and you'll have a reasonably accurate sketch.

      If you also need to multiply it out, deal with the complex roots first, and then just continue distributing until you're done.
      (3 votes)
  • male robot hal style avatar for user Khan
    The first question is incorrect because if it is a positive leading coefficient and an odd degree it is supposed to go up on the right side and down on the left side.
    (3 votes)
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  • male robot donald style avatar for user Carter Namkung
    in the last question, choice B fits in the explanation for the solution. It crosses (2,0) and (-1,0). If so, why is it not one of the answers?
    (3 votes)
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  • old spice man blue style avatar for user Jonathan Klokus
    I am really confused on this lesson in general. Is there a video i can watch somewhere?
    (2 votes)
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  • mr pants purple style avatar for user Mog Mogg
    The first question's answer has 2 real roots but that polynomial's highest degree is 3 which means it should have ether 1 real + 2 imaginary roots or 3 real roots. How is that possible?
    (0 votes)
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    • piceratops ultimate style avatar for user Fai
      It has 3 real roots, but the catch is one root has a multiplicity of 2. For example, this polynomial would be something like
      -1/2 x^3 + 3x^2 -9/2x + 2

      Which can be factored into something like:
      y=-1/2 (x-4)(x-1)^2
      (3 votes)
  • mr pants teal style avatar for user Janine Sibayan
    For #3, do we have to assume that "m" is the highest degree? I realise that this is how you would organise an equation for simplicity, but it could be taken as a trick question if "n" were to be the highest degree.
    (1 vote)
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    • starky ultimate style avatar for user Graham Earle
      It doesn't matter whether the "n" or "m" is higher because they've told us whether they're even or odd. Even degree roots will "bounce" off of the x-axis and odd degree roots will cross it but it doesn't matter which number is larger if all we're looking at is the behavior of the function at its roots.
      (1 vote)
  • aqualine seed style avatar for user Bell Crate
    In problem #1, i found the end behavior of ax^3 + bx^2 + cx + 2. I found that as x approaches +infinity, y approaches +infinity and as x approaches -infinity, y approaches -infinity. This is because of ax^3. a>0 and the exponent is an odd number, therefore, as x approaches +infinity, y approaches +infinity and as x approaches -infinity, y approaches -infinity. Problem #1 shows the exact opposite. When I clicked into [i need help] it just explained why C was correct because of the y intercept, and that they couldn't determine end behavior because they don't know the specific values of a, b, and c. If the end behavior is as they show in problem #1, wouldn't the equation look like -ax^3 + bx^2 + cx + 2?
    (1 vote)
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    • stelly blue style avatar for user Kim Seidel
      The problem tells you that "a", "b" and "c" are real numbers. This means they can be either positive or negative. You don't know which. So, you can't determine the end-behavior.
      The only real feature you can determine is that the Y-intercept = (0, 2). The only graph with that point it option C.
      (1 vote)
  • piceratops seed style avatar for user mancat18
    When I graphed number 3 in my calculator the graph looked just like D. How could A be an answer as well? Does it have do do with the numbers that you insert?
    (1 vote)
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