Describe the end behavior of f(x)=-x^5+4x^3-2x-2

As x→∞, f(x)→∞

and as x→-∞, f(x)→-∞

Step-by-step explanation:

View Image.

Your first step is to look at the highest value exponent.

In this case it's x⁵.

If the highest exponent is an odd number, the graph will look like a sideway S as shown in the picture below. You can see as you move farther to the right, the y value approaches positive  infinity hence "As x→∞, f(x)→∞".

As you move farther left, the graph reaches negative infinity, hence "as x→-∞, f(x)→-∞".

Anyway, if your exponent were to be an even number then you're graph would look like a parabola (looks like a letter U) and f(x) would reach positive infinity when moving to the right side and left side.

it is true

step-by-step explanation:

all the combinations add up i used   calculator

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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