Question

find (a) the leading term of the polynomial, (b) the limit as x approaches ∞, and (c) the limit as x approaches -∞. p(x)=16 + 4x² - 8x³ (a) the leading term is

Explanation:

Step1: Identify leading term

The leading term of a polynomial is the term with the highest - degree. In the polynomial \(p(x)=16 + 4x^{2}-8x^{3}\), the degree of the first term \(16\) (or \(16x^{0}\)) is \(0\), the degree of the second term \(4x^{2}\) is \(2\), and the degree of the third term \(-8x^{3}\) is \(3\). Since \(3>2 > 0\), the leading term is \(-8x^{3}\).

Step2: Find limit as \(x

ightarrow\infty\)
We know that for a polynomial \(y = a_{n}x^{n}+a_{n - 1}x^{n-1}+\cdots+a_{0}\), \(\lim_{x
ightarrow\infty}y\) is determined by the leading - term \(a_{n}x^{n}\). Here \(n = 3\) (odd) and \(a_{n}=-8<0\). So \(\lim_{x
ightarrow\infty}(-8x^{3})=-\infty\).

Step3: Find limit as \(x

ightarrow-\infty\)
For the leading term \(y=-8x^{3}\), when \(x
ightarrow-\infty\), we have \(\lim_{x
ightarrow-\infty}(-8x^{3})=\infty\) because when we substitute a very large negative number \(-M\) (\(M>0\)) into \(-8x^{3}\), we get \(-8(-M)^{3}=8M^{3}\) which goes to \(\infty\) as \(M
ightarrow\infty\).

Answer:

(A) \(-8x^{3}\)
(B) \(-\infty\)
(C) \(\infty\)