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.
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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 y-axis and down on the right side of the y-axis.
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.
I'm really confused now .... They represent all...
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)
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 y-axis and negative on the other. Multiplying by a negative value for a flips the diagram over along the y-axis, changing up/down to down/up and vice versa.
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.
Remember, you only look at the first term — plugging x=1 into the whole thing won't necessarily give you the right answer!
ok so what is the right answer? :)
I gave it to you in my first post: The end behavior is up on the left side of the y-axis and down on the right side of the y-axis.