What is the end behavior for f(x)=(2x−1)^2 (2−x)

Answer 1

What you want is to explain is

As x approaches -∞ then f(-∞) approaches ??

As x approaches ∞ then f(∞) approaches ??

This can be determined by showing the graph of the equation or by expanding the polynomial and observing the sign of the coefficient of the leading term.

Answer 2

For polynomials the end-behavior is controlled by the term with the highest exponent. You do not have to actually multiply this all out to get the polynomial in standard form. If you square the binomial perhaps you can see the leading term will be 4x². If you then multiply that perfect square trinomial by (2 - x), when you distribute the -x the leading term will become -4x³. As x approaches negative infinity the x³ factor is also approaching negative infinity but the minus sign in front results in an end-behavior of positive infinity as x -> -∞. Similarly as x ->∞, the x³ factor approaches ∞, but the -4 factor sends the y values to -∞. The end-behavior is up-down.

(2x-1)²(2 - x) = (4x²-4x+1)(2-x) = -4x³+4x²-x +8x²-8x+2= -4x³+12x²-9x+2

Check it out here(select the following URL, right-click, and choose "go to...)

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desmos.com/calculator/m30z6zovur

Answer 3

end behavior

y approaches- infinity and - infinity, as highest degreed term dominates as x gets very large or small

f(x) = (2x-1)^2(2-x)= (4x^2 -4x +1)(2-x)= -4x^3 +16x^2=9x+2

-4x^3 dominates and it approaches -infinity as x grows very large and approaches +infinity as x grows very small