Question

consider the following polynomial.

h(x) = -2x² + 4x - 11

step 2 of 2: describe the behavior of the graph of h(x) as x → ±∞.

answer 4 points

h(x) →
as x → -∞

h(x) →
as x → ∞

Explanation:

Step1: Identify the leading term

The polynomial \( h(x) = -2x^2 + 4x - 11 \) has a leading term of \( -2x^2 \). The degree of the polynomial is 2 (even) and the leading coefficient is -2 (negative).

Step2: Analyze end - behavior for even - degree polynomials

For a polynomial with an even degree, if the leading coefficient is negative, as \( x\to\pm\infty \), the value of the polynomial will tend to \( -\infty \). This is because when we take \( x\) to be a very large positive or negative number, the term with the highest power (the leading term) dominates the behavior of the polynomial. For \( y = ax^n\) where \( n\) is even:

• If \( a>0 \), as \( x\to\pm\infty \), \( y\to\infty \)
• If \( a < 0 \), as \( x\to\pm\infty \), \( y\to-\infty \)

In our case, \( n = 2\) (even) and \( a=-2<0 \). So when \( x\to-\infty \), \( h(x)\to-\infty \) and when \( x\to\infty \), \( h(x)\to-\infty \)

Answer:

For \( h(x)\to\) \(\boldsymbol{-\infty}\) as \( x\to-\infty\) and \( h(x)\to\) \(\boldsymbol{-\infty}\) as \( x\to\infty\)