Question

Which of the following graphs represents the function f(x) = -2x^3 - x^2 + 3x + 1

Answer 1

Welcome to Open Study! Having any questions or problems? Just ask me or anyone with a purple A next to their username!

Answer 2

The end behavior of this function will be very much like that of -x^3.
As x -> - inf, y-> +inf and as x -> + inf, y-> -inf

Answer 3

I still don't understand lol

Answer 4

This is a cubic function. The last two graphs are more or less like a parabola and so you can rule them out. It is either the first or the second choice.
Put x = 0, what is f(x)? How about x = -1 and x = +1.
You can quickly find out which graph it is.

f(x) = -2x^3 - x^2 + 3x + 1
f(0) = ?
f(1) = ?
f(-1) = ?

Answer 5

Oh i understand what your saying now

Answer 6

cool. what is your answer?

Answer 7

Wait so f(1) = -2x^3 - x^2 + 3x + 1 is -3 ?
and f(-1) = = -2x^3 - x^2 + 3x + 1 is 7 ?
so would the answer be B?

Answer 8

f(x) = -2x^3 - x^2 + 3x + 1
f(0) = 0 0 0 + 1 = 1
f(1) = -2(1)^3 - (1)^2 + 3(1) + 1 = -2 - 1 + 3 + 1 = 1
f(-1) = -2(-1)^3 - (-1)^2 + 3(-1) + 1 = +2 - 1 - 3 + 1 = -1

Answer 9

OOOOOh wow i'm so bad at math im so sorry, it would be A then correct because it touches at (1,0) ?

Answer 10

We are looking for a curve that has the points: (0, 1), (1, 1) and (-1, -1)

Answer 11

X Y
0 1
1 1
-1 -1

Answer 12

Oh okay I see what your saying, so it's B. so whenever I do problems like these I just look for the points on the graph the curves make? One more question, how do I know which number to plug into f(x) or is it just any numbers?

Btw thanks a lot for helping me because math is my worst subject.

Answer 13

B is correct.
x = 0, 1 and -1 are the EASIEST points to plug into a polynomial.
In this problem, for example, f(x) = -2x^3 - x^2 + 3x + 1
It is so easy to find f(0). All the x terms will be 0 and you will be left with 1.
So when x = 0, f(0) = 1 (that is, when x = 0, y = 1 on the graph).
This one point alone is sufficient to rule out A, C and D and you can correctly conclude it is B.

Answer 14

Thanks a bunch you really helped me a lot!

Answer 15

You are very welcome.