End Behavior and Local Behavior of Rational Function Practice

The end behavior of a function $f$ describes the trend of the function at the "ends" of the $x$-axis, i.e., as $x$ approaches $+\infty$ and as $x$ approaches $-\infty$.

The end behavior of any polynomial is similar to the end behavior of the polynomial that only contains the leading term.

For rational functions, the end behavior will be similar to the leading term in the numerator divided by leading term in the denominator.

In our case:

$\dfrac{ {-8x^4}+3x^2-7x}{ {4x^2}-4}\approx\dfrac{ {-8x^4}}{ {4x^2}}$

We can now reduce $\dfrac{-8x^4}{4x^2}$ to lowest terms:

$\dfrac{-8x^4}{4x^2} =\dfrac{-2\cdot \not 4\cdot \not {x^2}\cdot x^2}{ \not 4\cdot \not {x^2}}=-2x^2$

What does $-2x^2$ approach zs $x$ approaches
$\infty$ and $-\infty$?

As $x$ gets larger in the negative direction, the value of $-2x^2$ gets larger in the negative direction. We write this as:

$-2x^2\to -\infty$ as $x\to -\infty$.

As $x$ gets larger in the positive direction, the value of $-2x^2$ gets larger in the negative direction. We write this as:

$-2x^2\to -\infty$ as $x\to \infty$.

$f$ should behave the same.

This is $f$'s end behavior:

$f(x)\to -\infty$ as $x\to -\infty$.

$f(x)\to -\infty$ as $x\to \infty$.

The end behavior of a function $f$ describes the trend of the function at the "ends" of the $x$-axis, i.e., as $x$ approaches $+\infty$ and as $x$ approaches $-\infty$.

The end behavior of any polynomial is similar to the end behavior of the polynomial that only contains the leading term.

For rational functions, the end behavior will be similar to the leading term in the numerator divided by leading term in the denominator.

In our case:

$\dfrac{ {-8x^3}-3x+4}{ {2x^2}-9}\approx\dfrac{ {-8x^3}}{ {2x^2}}$

We can now reduce $\dfrac{-8x^3}{2x^2}$ to lowest terms:

$\dfrac{-8x^3}{2x^2} =\dfrac{-4\cdot \not 2\cdot \not{x^2}\cdot x}{ \not 2\cdot \not{x^2}}=-4x$

What does $-4x$ approach as $x$ approaches $\infty$ and $-\infty$?

As $x$ gets larger in the negative direction, the value of $-4x$ gets larger in the positive direction. We write this as:

$-4x\to \infty$ as $x\to -\infty$.

As $x$ gets larger in the positive direction, the value of $-4x$ gets larger in the negative direction. We write this as:

$-4x\to -\infty$ as $x\to \infty$.

$f$ should behave the same.

This is $f$'s end behavior:

$f(x)\to \infty$ as $x\to -\infty$.

$f(x)\to -\infty$ as $x\to \infty$.

The end behavior of a function $f$ describes the trend of the function at the "ends" of the $x$-axis, i.e., as
$x$ approaches $+\infty$ and as $x$ approaches $-\infty$.

The end behavior of any polynomial is similar to the end behavior of the polynomial that only contains the leading term.

For rational functions, the end behavior will be similar to the leading term in the numerator divided by leading term in the denominator.

In our case:

$\dfrac{ {-8x^3}+5x^2-1}{ {2x^3}-9x}\approx\dfrac{ {-8x^3}}{ {2x^3}}$

We can now reduce $\dfrac{-8x^3}{2x^3}$ to lowest terms:

$\dfrac{-8x^3}{2x^3} =\dfrac{-4\cdot \not{2}\cdot \not{x^3}}{ \not{2}\cdot \not{x^3}}=-4$

This means that as $x$ approaches $\infty$ and $-\infty$, our function $f$ behaves like the constant function $y=-4$.

Specifically, as $x$ gets larger in the negative direction, $f$ gets closer and closer to $-4$. We write this as:$f(x)\to -4$ as $x\to -\infty$.

As $x$ gets larger in the positive direction, $f$ gets closer and closer to $-4$. We write this as:

$f(x)\to -4$ as $x\to \infty$.

This is $f$'s end behavior:

$f(x)\to -4$ as $x\to -\infty$.

$f(x)\to -4$ as $x\to \infty$.

The end behavior of a function $f$ describes the trend of the function at the "ends" of the $x$-axis, i.e., as $x$ approaches $+\infty$ and as $x$ approaches $-\infty$.

The end behavior of any polynomial is similar to the end behavior of the polynomial that only contains the leading term.

For rational functions, the end behavior will be similar to the leading term in the numerator divided by leading term in the denominator.

In our case:

$\dfrac{ {-4x^3}+7x+9}{ {8x^6}-9x^4-2x}\approx\dfrac{ {-4x^3}}{ {8x^6}}$

We can now reduce $\dfrac{-4x^3}{8x^6}$ to lowest terms:

$\dfrac{-4x^3}{8x^6} =\dfrac{-1\cdot \not{4}\cdot \not{x^3}}{2\cdot \not{4}\cdot \not{x^3}\cdot x^3}=-\dfrac{1}{2x^3}$

What does $-\dfrac{1}{2x^3}$ approach as $x$ approaches $\infty$ and $-\infty$?

As $x$ gets larger in the negative direction, the value of $-\dfrac{1}{2x^3}$ gets closer and closer to zero. We write this as:

$-\dfrac{1}{2x^3}\to 0$ as $x\to -\infty$.

As $x$ gets larger in the positive direction, the value of $-\dfrac{1}{2x^3}$ gets closer and closer to zero. We write this as:

$-\dfrac{1}{2x^3}\to 0$ as $x\to \infty$. 

$f$ should behave the same.

This is $f$'s end behavior:

$f(x)\to 0$ as $x\to -\infty$.

$f(x)\to 0$ as $x\to \infty$.