Question

§4.5: use arrow notation to describe the end behavior of the function. 8.) $g(x) = -\frac{2x^2 + 5x}{11x^2 - 3x + 16}$

Explanation:

Step1: Identify the degrees of numerator and denominator

The numerator is $-2x^2 + 5x$ (we can consider the leading term $-2x^2$), so the degree of the numerator, $n = 2$. The denominator is $11x^2 - 3x + 16$, and its leading term is $11x^2$, so the degree of the denominator, $m = 2$. Since $n = m$, we look at the ratio of the leading coefficients.

Step2: Find the ratio of leading coefficients

The leading coefficient of the numerator (from the leading term $-2x^2$) is $-2$, and the leading coefficient of the denominator (from the leading term $11x^2$) is $11$. The ratio of these leading coefficients is $\frac{-2}{11}$.

Step3: Determine the end - behavior

For a rational function $f(x)=\frac{a_nx^n+\cdots}{b_mx^m+\cdots}$, when $n = m$, as $x
ightarrow\pm\infty$, $f(x)
ightarrow\frac{a_n}{b_m}$. So for $g(x)=-\frac{2x^2 + 5x}{11x^2-3x + 16}$, as $x
ightarrow\infty$, we consider the limit $\lim_{x
ightarrow\infty}-\frac{2x^2+5x}{11x^2 - 3x + 16}$. Dividing both the numerator and denominator by $x^2$ (since $x
ightarrow\infty$, $x
eq0$), we get $\lim_{x
ightarrow\infty}-\frac{2+\frac{5}{x}}{11-\frac{3}{x}+\frac{16}{x^2}}$. As $x
ightarrow\infty$, $\frac{5}{x}
ightarrow0$, $\frac{3}{x}
ightarrow0$, and $\frac{16}{x^2}
ightarrow0$. So $\lim_{x
ightarrow\infty}g(x)=-\frac{2}{11}$. Similarly, as $x
ightarrow-\infty$, $\lim_{x
ightarrow-\infty}-\frac{2x^2 + 5x}{11x^2-3x + 16}$. Dividing numerator and denominator by $x^2$ (note that $x^2>0$ when $x
ightarrow-\infty$), we have $\lim_{x
ightarrow-\infty}-\frac{2+\frac{5}{x}}{11-\frac{3}{x}+\frac{16}{x^2}}$. Again, as $x
ightarrow-\infty$, $\frac{5}{x}
ightarrow0$, $\frac{3}{x}
ightarrow0$, and $\frac{16}{x^2}
ightarrow0$. So $\lim_{x
ightarrow-\infty}g(x)=-\frac{2}{11}$. In arrow notation, this means as $x
ightarrow\infty$, $g(x)
ightarrow-\frac{2}{11}$ and as $x
ightarrow-\infty$, $g(x)
ightarrow-\frac{2}{11}$.

Answer:

As $x
ightarrow\infty$, $g(x)
ightarrow-\frac{2}{11}$ and as $x
ightarrow-\infty$, $g(x)
ightarrow-\frac{2}{11}$ (or in arrow notation: $x
ightarrow\infty\Rightarrow g(x)
ightarrow-\frac{2}{11}$ and $x
ightarrow-\infty\Rightarrow g(x)
ightarrow-\frac{2}{11}$)