Question

question 8 of 10 for the polynomial f(x) = 2x³ - 9x² + 11x - 20 as x → -∞, f(x) → ∞ a. true b. false

Explanation:

Step1: Recall End Behavior Rule

For a polynomial \( f(x)=a_nx^n + a_{n - 1}x^{n-1}+\dots+a_1x + a_0 \), the end - behavior is determined by the leading term \( a_nx^n \), where \( n \) is the degree of the polynomial and \( a_n \) is the leading coefficient. The degree \( n \) and the sign of \( a_n \) determine the end - behavior:

• If \( n \) is odd:
• If \( a_n>0 \), as \( x

ightarrow\infty \), \( f(x)
ightarrow\infty \); as \( x
ightarrow-\infty \), \( f(x)
ightarrow-\infty \).

• If \( a_n < 0 \), as \( x

ightarrow\infty \), \( f(x)
ightarrow-\infty \); as \( x
ightarrow-\infty \), \( f(x)
ightarrow\infty \).

• If \( n \) is even:
• If \( a_n>0 \), as \( x

ightarrow\pm\infty \), \( f(x)
ightarrow\infty \).

• If \( a_n < 0 \), as \( x

ightarrow\pm\infty \), \( f(x)
ightarrow-\infty \).

Step2: Identify Leading Term

For the polynomial \( f(x)=2x^3-9x^2 + 11x-20 \), the leading term is \( 2x^3 \). Here, the degree \( n = 3 \) (which is odd) and the leading coefficient \( a_n=2>0 \).

Step3: Determine End Behavior

Since \( n = 3 \) (odd) and \( a_n = 2>0 \), when \( x
ightarrow-\infty \), we substitute \( x \) with a very large negative number in the leading term \( 2x^3 \). Let \( x=-k \) where \( k
ightarrow\infty \). Then \( 2x^3=2(-k)^3=- 2k^3 \). As \( k
ightarrow\infty \), \( -2k^3
ightarrow-\infty \). So as \( x
ightarrow-\infty \), \( f(x)
ightarrow-\infty \), not \( \infty \).

Answer:

B. False