Solve and Graph Rational Inequalities - Pre-Calculus

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Determine the zeros of the following function and the points where the function will be undefined.

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The zeros of the function are the values of where the function will be equal to zero. In order to find these we set the numerator of the function equal to zero.

We only need to solve for once,

So the zeros of this function are .

To solve for the points at which this function will be undefined, we set the denominator equal to zero and solve for .

$x= \pm 5$

And so the function is undefined at $x= \pm 5$

Solve and graph:

$$\frac{4x + 3}{4}$ - $\frac{2x}{x + 1}$ < x$

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Multiply both sides of the equation by the common denominator of the fractions:

$4(x + 1)($\frac{4x + 3}{4}$ - $\frac{2x}{x + 1}$) < x\cdot4(x + 1)$

Simplify:

$$\frac{3}{5}$ < x$

For standard notation, and the fact that inequalities can be read backwards:

$x > $\frac{3}{5}$$

For interval notation:

$($\frac{3}{5}$, +\infty)$

Graph:

Solve and graph:

$$\frac{3}{x - 2}$ < 1$

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Graph the rational expression,

$y = $\frac{3}{x - 2}$$

Because and a divide by is undefined in the real number system, there is a vertical asymptote where .

As $\rightarrow$ , $\rightarrow$ $+\infty$ , and as $\rightarrow +\infty$ , $\rightarrow 0$ .

As $\rightarrow $2^{-}$$ , $\rightarrow -\infty$ , and as $\rightarrow$ $-\infty$ , $\rightarrow 0$ .

The funtion y is exists over the allowed x-intervals:

One approach for solving the inequality:

For $y = $\frac{3}{x - 2}$$

Determine where over the x-values or .

for the intervals $(-\infty, 2)$ or $(5, +\infty )$ .

Then the solution is $(-\infty, 2) \cup (5, +\infty )$ .

Another approach for solving the inequality:

Write $$\frac{3}{x - 2}$ < 1$ as $y = $\frac{3}{x - 2}$$ , then determine the x-values that cause to be true:

is true for or .

Then the solution is $(-\infty, 2) \cup (5, +\infty )$ .

Solve the inequality.

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First, subtract from both sides so you get

.

Then find the common denominator and simplify

.

Next, factor out the numerator

$$\frac{(3x-1)(x-2)}{x+1}$\leq0$

and set each of the three factor equal to zero and solve for .

The solutions are

$x=-1;$\frac{1}{3}$;2$ .

Now plug in values between $(-\infty ,-1)$ , $\left(-1,$\frac{1}{3}\right]$ , $\left[$\frac{1}{3}$,2\right]$ , and $[2,\infty)$ into the inequality and observe if the conditions of the inequality are met.

Note that $x\neq-1$ . They are met in the interval $(-\infty ,-1)$ and $\left[$\frac{1}{3}$,2\right]$ .

Thus, the solution to the inequality is

$(-\infty ,-1)\cup \left[$\frac{1}{3}$,2\right]$ .

Which of the following is a graph of the function:

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We begin by finding the zeros of the equation using the numerator.

So we know that the function will equal zero when . If we just look at the numerator of the function, then this graph would be a parabola with its point at . Now we will solve for the points where the function is undefined by setting the denominator equal to zero and solving for .

And so the function is undefined at . If we make a table to solve for some of the points of the graph:

x y
$-$\frac{4}{5}$$
$-$\frac{1}{6}$$
$-$\frac{1}{8}$$
$-$\frac{4}{9}$$
$\frac{144}{5}$
$\frac{169}{6}$
$\frac{225}{8}$
$\frac{246}{9}$

And if we graph these points we see something like below (which is our answer). Note that the dotted blue line is the vertical asymptote at .

Which of the following best describes the statement:

The undefined points of rational functions are vertical asymptotes.

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When solving for a point where the function will be undefined, you set the denominator equal to zero and solve for . This creates a vertical asymptote because when the denominator equals zero the function is undefined and we are solving for . Say for example a function is undefined at . So at all values of where this function is undefined creating a vertical asymptote.

True or False: For the inequality , is a potential value for .

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This inequality wants all values where is greater than . So everything up until is included and this is represented by having a dotted line on the graph or an open circle on a number line.

Which of the following is an accurate graph of the inequality ?

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We will first begin by solving for the zeros and undefined points of the inequality. We solve for the zeros by setting the numerator equal to zero.

$\left ( x+2 \right )\left ( x-1 \right )=0$

And so the zeros of this function are at

Now we will solve for the undefined points by setting the denominator equal to zero. Since the denominator is , then whenever , the function is undefined. Now we need to find for which values of is each factor is greater than zero.

For , any value where will be positive and we will be able to graph it. For , any value where will be positive. Now, we can only graph these values up until because the function is undefined here. We are able to pick up the graph again once we reach . The graph will look like the one below.

Determine the zeros of the following function and the points where the function will be undefined.

Tap to see back →

The zeros of the function are the values of where the function will be equal to zero. In order to find these we set the numerator of the function equal to zero.

We only need to solve for once,

So the zeros of this function are .

To solve for the points at which this function will be undefined, we set the denominator equal to zero and solve for .

$x= \pm 5$

And so the function is undefined at $x= \pm 5$

Solve and graph:

$$\frac{4x + 3}{4}$ - $\frac{2x}{x + 1}$ < x$

Tap to see back →

Multiply both sides of the equation by the common denominator of the fractions:

$4(x + 1)($\frac{4x + 3}{4}$ - $\frac{2x}{x + 1}$) < x\cdot4(x + 1)$

Simplify:

$$\frac{3}{5}$ < x$

For standard notation, and the fact that inequalities can be read backwards:

$x > $\frac{3}{5}$$

For interval notation:

$($\frac{3}{5}$, +\infty)$

Graph:

Solve and graph:

$$\frac{3}{x - 2}$ < 1$

Tap to see back →

Graph the rational expression,

$y = $\frac{3}{x - 2}$$

Because and a divide by is undefined in the real number system, there is a vertical asymptote where .

As $\rightarrow$ , $\rightarrow$ $+\infty$ , and as $\rightarrow +\infty$ , $\rightarrow 0$ .

As $\rightarrow $2^{-}$$ , $\rightarrow -\infty$ , and as $\rightarrow$ $-\infty$ , $\rightarrow 0$ .

The funtion y is exists over the allowed x-intervals:

One approach for solving the inequality:

For $y = $\frac{3}{x - 2}$$

Determine where over the x-values or .

for the intervals $(-\infty, 2)$ or $(5, +\infty )$ .

Then the solution is $(-\infty, 2) \cup (5, +\infty )$ .

Another approach for solving the inequality:

Write $$\frac{3}{x - 2}$ < 1$ as $y = $\frac{3}{x - 2}$$ , then determine the x-values that cause to be true:

is true for or .

Then the solution is $(-\infty, 2) \cup (5, +\infty )$ .

Solve the inequality.

Tap to see back →

First, subtract from both sides so you get

.

Then find the common denominator and simplify

.

Next, factor out the numerator

$$\frac{(3x-1)(x-2)}{x+1}$\leq0$

and set each of the three factor equal to zero and solve for .

The solutions are

$x=-1;$\frac{1}{3}$;2$ .

Now plug in values between $(-\infty ,-1)$ , $\left(-1,$\frac{1}{3}\right]$ , $\left[$\frac{1}{3}$,2\right]$ , and $[2,\infty)$ into the inequality and observe if the conditions of the inequality are met.

Note that $x\neq-1$ . They are met in the interval $(-\infty ,-1)$ and $\left[$\frac{1}{3}$,2\right]$ .

Thus, the solution to the inequality is

$(-\infty ,-1)\cup \left[$\frac{1}{3}$,2\right]$ .

Which of the following is a graph of the function:

Tap to see back →

We begin by finding the zeros of the equation using the numerator.

So we know that the function will equal zero when . If we just look at the numerator of the function, then this graph would be a parabola with its point at . Now we will solve for the points where the function is undefined by setting the denominator equal to zero and solving for .

And so the function is undefined at . If we make a table to solve for some of the points of the graph:

x y
$-$\frac{4}{5}$$
$-$\frac{1}{6}$$
$-$\frac{1}{8}$$
$-$\frac{4}{9}$$
$\frac{144}{5}$
$\frac{169}{6}$
$\frac{225}{8}$
$\frac{246}{9}$

And if we graph these points we see something like below (which is our answer). Note that the dotted blue line is the vertical asymptote at .

Which of the following best describes the statement:

The undefined points of rational functions are vertical asymptotes.

Tap to see back →

When solving for a point where the function will be undefined, you set the denominator equal to zero and solve for . This creates a vertical asymptote because when the denominator equals zero the function is undefined and we are solving for . Say for example a function is undefined at . So at all values of where this function is undefined creating a vertical asymptote.

True or False: For the inequality , is a potential value for .

Tap to see back →

This inequality wants all values where is greater than . So everything up until is included and this is represented by having a dotted line on the graph or an open circle on a number line.

Which of the following is an accurate graph of the inequality ?

Tap to see back →

We will first begin by solving for the zeros and undefined points of the inequality. We solve for the zeros by setting the numerator equal to zero.

$\left ( x+2 \right )\left ( x-1 \right )=0$

And so the zeros of this function are at

Now we will solve for the undefined points by setting the denominator equal to zero. Since the denominator is , then whenever , the function is undefined. Now we need to find for which values of is each factor is greater than zero.

For , any value where will be positive and we will be able to graph it. For , any value where will be positive. Now, we can only graph these values up until because the function is undefined here. We are able to pick up the graph again once we reach . The graph will look like the one below.

Solve and graph:

$$\frac{4x + 3}{4}$ - $\frac{2x}{x + 1}$ < x$

Tap to see back →

Multiply both sides of the equation by the common denominator of the fractions:

$4(x + 1)($\frac{4x + 3}{4}$ - $\frac{2x}{x + 1}$) < x\cdot4(x + 1)$

Simplify:

$$\frac{3}{5}$ < x$

For standard notation, and the fact that inequalities can be read backwards:

$x > $\frac{3}{5}$$

For interval notation:

$($\frac{3}{5}$, +\infty)$

Graph:

Solve and graph:

$$\frac{3}{x - 2}$ < 1$

Tap to see back →

Graph the rational expression,

$y = $\frac{3}{x - 2}$$

Because and a divide by is undefined in the real number system, there is a vertical asymptote where .

As $\rightarrow$ , $\rightarrow$ $+\infty$ , and as $\rightarrow +\infty$ , $\rightarrow 0$ .

As $\rightarrow $2^{-}$$ , $\rightarrow -\infty$ , and as $\rightarrow$ $-\infty$ , $\rightarrow 0$ .

The funtion y is exists over the allowed x-intervals:

One approach for solving the inequality:

For $y = $\frac{3}{x - 2}$$

Determine where over the x-values or .

for the intervals $(-\infty, 2)$ or $(5, +\infty )$ .

Then the solution is $(-\infty, 2) \cup (5, +\infty )$ .

Another approach for solving the inequality:

Write $$\frac{3}{x - 2}$ < 1$ as $y = $\frac{3}{x - 2}$$ , then determine the x-values that cause to be true:

is true for or .

Then the solution is $(-\infty, 2) \cup (5, +\infty )$ .

Solve the inequality.

Tap to see back →

First, subtract from both sides so you get

.

Then find the common denominator and simplify

.

Next, factor out the numerator

$$\frac{(3x-1)(x-2)}{x+1}$\leq0$

and set each of the three factor equal to zero and solve for .

The solutions are

$x=-1;$\frac{1}{3}$;2$ .

Now plug in values between $(-\infty ,-1)$ , $\left(-1,$\frac{1}{3}\right]$ , $\left[$\frac{1}{3}$,2\right]$ , and $[2,\infty)$ into the inequality and observe if the conditions of the inequality are met.

Note that $x\neq-1$ . They are met in the interval $(-\infty ,-1)$ and $\left[$\frac{1}{3}$,2\right]$ .

Thus, the solution to the inequality is

$(-\infty ,-1)\cup \left[$\frac{1}{3}$,2\right]$ .

Determine the zeros of the following function and the points where the function will be undefined.

Tap to see back →

The zeros of the function are the values of where the function will be equal to zero. In order to find these we set the numerator of the function equal to zero.

We only need to solve for once,

So the zeros of this function are .

To solve for the points at which this function will be undefined, we set the denominator equal to zero and solve for .

$x= \pm 5$

And so the function is undefined at $x= \pm 5$