Answer:
The functions that approach negative infinity are,
f(x) = (4x + 1)(3x+5)(x-2)
f(x)=-4.8x(2x+3)(x-9)(x+5)
f(x) = 0.2x(x+4)(x+7)(x+8)(x-2)
Explanation:
We only need to look at the ighest degree term to get an idea of the behaviour as x approaches infinity,
now,
for f(x) = (4x + 1)(3x+5)(x-2)
The highest degree term is x^3 (with some constant), and this goes to negative infinity as x goes to negative infinity.
f(x)=-4.8x(2x+3)(x-9)(x+5)
The highest degree term is -Cx^4 (with some constant C), this goes to negative infinity(i.e. x^4 goes to infinity so -x^4 goes to negative infinity) as x approaches infinity
f(x) = (4x+3)(x-5)(x+8)(x-3)
Here, the highest degree term x^4 has a positive coefficient so it approaches infinity as x approaches negative infinity
f(x) = -0.5x(3x-7)(4x + 1)(x+9)(x-3)
The highest degree is x^5, this goes to negative infinity but since thecoefficient is also negative (due to the negative sign at teh start), -x^5 approaches positive infinity as x approaches negative infinity
f(x) = 0.2x(x+4)(x+7)(x+8)(x-2)
The highest term is x^5 so it approaches negative infinity as x goes to negative infinity