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### Course: Integrated math 3 > Unit 4

Lesson 4: Putting it all together# 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!

## Want to join the conversation?

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Effort isn't fair. Effort isn't equal.*But effort always works!*(15 votes)

- Who else LOVES these questions?(26 votes)
**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)- 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)

- 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)
- It is a valid polynomial graph. Go to desmos.com and use their graphing calculator. Type in:

y=0.1(x+1)^3(x-3)^2

You will get a graph very similar to the one that you've commented on.(8 votes)

- 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)
- 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)

- I am really confused on this lesson in general. Is there a video i can watch somewhere?(5 votes)
- Rewatch the entire section in Algebra 2(5 votes)

- the graphing is confusing(5 votes)
- 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)
- In this case all the coefficients are real numbers, so they don't have to be positive.(7 votes)

- 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)- 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)

- I don't understand, why must it always go through -1?(3 votes)
- 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)