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## Algebra 2

### Unit 5: 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 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?

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?

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?

## Want to join the conversation?

• 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.
• 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.
• 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?
• 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
• 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.
• 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.
• 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.
• In this case all the coefficients are real numbers, so they don't have to be positive.
• 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?
• If you read the explanation more careful, you will see B is not the answer. Because B cuts the x-axis at 2, not touches, as it suppose to; and the curve touches x-axis at -1, not cuts as it suppose to.
• I am really confused on this lesson in general. Is there a video i can watch somewhere?
• Rewatch the entire section in Algebra 2
(1 vote)
• 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?
• 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