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Falkqwer
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I really need help with this equation PLEASE help if you can ^_^
Consider the leading term of the polynomial function. What is the end behavior of the graph?
–2x^7 + 8x^6 + 5x^5 + 4
a.)The leading term is –2x^7. Since n is odd, and a is negative, the end behavior is up and down.
b.)The leading term is –2x^7. Since n is odd, and a is negative, the end behavior is up and up.
c.)The leading term is –2x^7. Since n is odd, and a is negative, the end behavior is down and down.
d.)The leading term is –2x^7. Since n is odd, and a is negative, the end behavior is down and up.
 one year ago
 one year ago
Falkqwer Group Title
I really need help with this equation PLEASE help if you can ^_^ Consider the leading term of the polynomial function. What is the end behavior of the graph? –2x^7 + 8x^6 + 5x^5 + 4 a.)The leading term is –2x^7. Since n is odd, and a is negative, the end behavior is up and down. b.)The leading term is –2x^7. Since n is odd, and a is negative, the end behavior is up and up. c.)The leading term is –2x^7. Since n is odd, and a is negative, the end behavior is down and down. d.)The leading term is –2x^7. Since n is odd, and a is negative, the end behavior is down and up.
 one year ago
 one year ago

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Falkqwer Group TitleBest ResponseYou've already chosen the best response.0
So its up up right?
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.1
The leading term is 2x^7, with a = 2, and n = 7. Let's consider what happens as x goes from 0 to a large positive number: x^7 gets really large, and is multiplied by a negative number, so y becomes a really large negative number as x increases. That sounds like "down" to me. Now consider what happens as x goes from 0 to a large negative number: because n is odd, and the product of two negative numbers is a positive number, but the product of a negative number and a positive number is a negative number, x^7 is negative when x < 0. Now we are multiplying it by another negative number, so ax^7 is positive when a < 0 and x < 0, and it increases rapidly as x decreases (>  infinity). That sounds like "up" to me. The end behavior is up on the left side of the yaxis and down on the right side of the yaxis. I've attached 4 pictures showing the respective graphs of 2x^7, 2x^7, 2x^6, 2x^6 so you can see how the values of a and n affect the shape of the curve.
 one year ago

Falkqwer Group TitleBest ResponseYou've already chosen the best response.0
I'm really confused now .... They represent all...
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.1
Sorry, it changed my file names :( First graph is up/down, and a = 2, n = 7 (2x^7) Second graph is down/up, and a = 2, n = 7 (2x^7) Third graph is down/down, and a = 2, n = 6 (2x^6) Fourth graph is up/up, and a = 2, n = 6 (2x^6)
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.1
As the value of x goes far away from 0, only the leading term matters, because it has a greater exponent for x than any of the others, and eventually that extra power of x will outweigh even the largest coefficient that might be on the lower power terms. If we have an even power of x (n is even), x^n is always positive. 1*1=1, (1)*(1)=1. That is a bowl opening upward, and if we multiply it by a negative value for a, we just flip it over. Do you agree with that? If we have an odd power of x (n is odd), x^n is positive on one side of the yaxis and negative on the other. Multiplying by a negative value for a flips the diagram over along the yaxis, changing up/down to down/up and vice versa.
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.1
To figure out whether it is up/down or down/up with an odd power, just plug in x = 1. ax^n = a(1)^n = a. If a is positive, then it is going up on the +x side and down on the x side.
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.1
Remember, you only look at the first term — plugging x=1 into the whole thing won't necessarily give you the right answer!
 one year ago

Falkqwer Group TitleBest ResponseYou've already chosen the best response.0
ok so what is the right answer? :)
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.1
I gave it to you in my first post: The end behavior is up on the left side of the yaxis and down on the right side of the yaxis.
 one year ago
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