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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=ax3+bx2+cx+2, where a, b, and c are real numbers?
Choose 1 answer:

2*) Which of the following could be the graph of y=2x5+p(x), where p(x) is a fourth degree polynomial?
Choose 1 answer:

3*) Which of the following could be the graph of y=k(x2)m(x+1)n, where k is a real number, m is an even integer, and n is an odd integer?
Choose all answers that apply:

Want to join the conversation?

  • starky tree style avatar for user 211202
    i hate maths :3
    (24 votes)
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    • leaf green style avatar for user pateyc8
      If you struggle to force your brain to process things you aren't interested in, then I encourage you to consider asking your parents about consulting an ADHD specialist or general psychiatrist. If diagnosed early enough, ADHD meds and therapy can help you to outgrow ADHD completely by the time you're an adult. The most important thing is to build good habits. Adderall basically saved my life. My only regret is getting diagnosed too late, so I feel compelled to remind people of the option.

      As an aside: identifying as "someone who hates math" is like deliberately setting the difficulty to the highest level. It is very important to be aware of how we choose to identify ourselves. We can't completely control our self-concept, but we will inevitably believe what we tell ourselves. Fake it 'til you make it! Choose to identify with habits you can admire, and try to build a reputation on those good habits. Even better if you make friends with people who already have those habits. In other words, use peer pressure to your advantage! "People know me as someone who gets all their schoolwork done." It makes it way easier to motivate yourself.

      I'm in college for psychology. I'm here because I wish to be "someone who's good at math," partially because math literacy directly correlates to higher income lol. tbh, I'm way behind my peers and I simply have no talent in math, but that doesn't matter! If I try and practice, I will improve!

      Effort isn't fair. Effort isn't equal. But effort always works!
      (15 votes)
  • leaf green style avatar for user Chiara  Menolascino
    Who else LOVES these questions?
    (26 votes)
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  • 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.
    (4 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.
      (13 votes)
  • duskpin ultimate style avatar for user C4LOwenZ
    I know there are an infinite array of polynomial equations, but I'll never believe there is a specific one to look like someone scribbling all over the graph paper, like Choice D in Question 3.
    (7 votes)
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  • 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?
    (8 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
      (5 votes)
  • 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?
    (5 votes)
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  • old spice man blue style avatar for user Yexziel Rivera
    the graphing is confusing
    (5 votes)
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  • eggleston blue 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.
    (2 votes)
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  • 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)
  • blobby green style avatar for user joychoi1009
    I don't understand, why must it always go through -1?
    (3 votes)
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    • stelly blue style avatar for user Kim Seidel
      I assume you are asking about problem 3. The polynomial is in factored form. If you set y=0 to find the x-intercepts, each factor gets set to 0 (using the zero product rule).
      x-2=0 and x+1=0
      Solve them, and you get that the x-intercepts are at x=-2 and x=-1

      Next, you look at the exponents on each factor. If the exponent is even, then the graph will bounce off the x-axis (or just touch it). If the exponent is odd, then the graph will cross over the x-axis.
      -- For the factor (x-2), the problem tells you that the exponent is even. So the graph will bounce off the x-axis at x=2
      -- For the factor (x+1), the problem tells you that the exponent is odd, so the graph will cross over the x-axis.

      Hope this helps.
      (4 votes)