How do you find the [katex] \textbf{x} [/katex]-intercept and the [katex] \textbf{y} [/katex]-intercept of rational functions?

You find the [katex] x [/katex]-intercepts and the [katex] y [/katex]-intercepts of rational functions applying the same strategy as you would with any function. For the [katex] x [/katex]-intercept set the function equal to [katex] 0 [/katex] and find the [katex] x [/katex] values. For the [katex] y [/katex]-intercept substitute [katex] 0 [/katex] in for the [katex] x [/katex] values and find [katex] y. [/katex] The [katex] x [/katex] and [katex] y [/katex] intercepts are real numbers.

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Rational functions

Rational function

Here you will learn about rational functions, including how to simplify rational functions and operations with rational functions.

Students first learn about rational functions in Algebra $1$ and expand their knowledge as they progress through high school math.

What are rational functions?

Rational functions or algebraic fractions, are functions that are fractions because they have a numerator and denominator. Rational functions are expressed as the ratio of two polynomials such that the denominator is not equal to $0.$

$R(x)=\cfrac{p(x)}{q(x)}$ where $q(x)$ ≠ $0.$

Rational functions contain at least one variable. They also have denominators and numerators that can be linear functions, quadratic functions, or any polynomial function.

Here are the three types of rational functions and the graph of the functions

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Simplifying rational functions is a necessary skill when applying operations to rational functions.

Let’s look at a few examples of how to simplify rational functions.

Remember when you learned how to simplify fractions in elementary school?

For example, if you have the fraction, $\cfrac{7}{21},$ you look to see if it can be simplified by determining if there are any common factors of the numerator and the denominator.

In this case, there is a common factor of $7$ for both the numerator and denominator.

\cfrac{7}{21}=\cfrac{7}{7 \times 3}=\cfrac{1}{3}

In order to simplify algebraic fractions or rational functions you will use a similar strategy. Look for a common factor to the numerator and the denominator.

Let’s simplify:

\cfrac{15 x^2}{30 x}

First, find the common factor between the numerator and the denominator.

Looking at the factors for both expressions, $3, 5,$ and $x$ are the common factors.

3 \times 5 \times x=15 x

$15x$ is the common factor.

If you want to learn more on how to simplify rational functions, go to:

Step-by-step guide: Simplifying rational expressions

Adding and subtracting rational expressions

Rational expressions can be added and subtracted. Similar to adding and subtracting fractions, rational expressions need a common denominator before they can be added or subtracted.

Let’s add the rational functions:

\cfrac{4}{24 x y}+\cfrac{10}{6 y^2}

Before the expressions can be added together, first you will need to find the common denominator.

The two denominators are:

24 x y \quad 6 y^2

Looking at the coefficients, the least common multiple of $24$ and $6$ is $24.$

To find the least common multiple of the variables, $xy$ and $y^2,$ you need to look at each variable and take the variable to the higher exponent.

$x$ has to be included in the least common denominator because it is the variable to the highest exponent $(x^1$ has a higher exponent than $x^0)$

\text { Remember } x^0=1 \text { meaning that } 6 y^2 \text { has } x^0

$y^2$ has to be included in the least common denominator because $y^2$ has a higher exponent than $y^1.$

So the least common denominator of $\cfrac{4}{24 x y}+\cfrac{10}{6y^2}$ is $24 x y^2.$

Now, you will have to adjust each rational expression so that it has a denominator of $24 x y^2.$ Be sure to multiply the numerator and denominator by the missing factors.

\begin{aligned}& \cfrac{4}{24 x y} \times \cfrac{y}{y}=\cfrac{4 y}{24 x y^2} \\\\ & \cfrac{10}{6 y^2} \times \cfrac{4 x}{4 x}=\cfrac{40 x}{24 x y^2} \end{aligned}

After adjusting both rational expressions to have a common denominator, you can add them.

$\cfrac{4 y}{24 x y^2}+\cfrac{40 x}{24 x y^2}=\cfrac{4 y+40 x}{24 x y^2} \rightarrow$ The rational expression can be written as one expression.

The numerators cannot be simplified any further because they are not like terms.

If you want to learn more about adding and subtracting rational functions, go to:

Step-by-step guide: Adding and subtracting rational expressions

Multiplying and dividing rational expressions

Rational expressions can be multiplied and divided. Similar to multiplying and dividing fractions, rational expressions do not need a common denominator to be multiplied or divided. The product or quotient should be fully simplified.

Let’s multiply the rational expressions.

\cfrac{(x-4)}{\left(x^2+8 x+15\right)} \times \cfrac{(x+5)^2}{\left(x^2-16\right)}

Before multiplying, factor any expression that can be factored.

In this case, the denominators of both expressions can be factored.

\begin{aligned}& \cfrac{(x-4)}{\left(x^2+8 x+15\right)} \times \cfrac{(x+5)^2}{\left(x^2-16\right)} \\\\ & \cfrac{(x-4)}{(x+3)(x+5)} \times \cfrac{(x+5)^2}{(x-4)(x+4)} \\\\ & \cfrac{(x-4)}{(x+3)(x+5)} \times \cfrac{(x+5)(x+5)}{(x-4)(x+4)} \end{aligned}

When multiplying or dividing, look to cancel any matching expressions as long as they are placed in opposite positions, (one in the numerator and one in the denominator)

The rational expressions can now be multiplied, meaning they can be written as one rational expression.

\cfrac{(x-4)(x+5)}{(x+3)(x+4)}=\cfrac{x^2+x-20}{x^2+7 x+12}

If you want to learn more about how to multiply and divide rational expressions, go to:

Step-by-step guide: Multiplying rational expressions

Solving rational equations

The strategy to solve rational equations is similar to the strategy used to solve equations with fractions in them.

Let’s solve the rational equation,

\cfrac{1}{x}+\cfrac{3}{2 x}=6

First, find the common denominator, in this case, the common denominator is $2x.$

Multiply the equation by the common denominator, $2x.$

\begin{aligned}& 2 x\left(\cfrac{1}{x}+\cfrac{3}{2 x}=6\right) \\\\ & 2 x \times \cfrac{1}{x}=\cfrac{2 x}{x}=2 \\\\ & 2 x \times \cfrac{3}{2 x}=\cfrac{6 x}{2 x}=3 \\\\ & 2 x \times 6=12 x \end{aligned}

The new equation after multiplying by $2x$ is:

2+3=12 x

Solving for $x\text{:}$

\begin{aligned}& 2+3=12 x \\\\ & 5=12 x \\\\ & \cfrac{5}{12}=\cfrac{12 x}{12} \\\\ & \cfrac{5}{12}=x \end{aligned}

Check the solution to make sure that $0$ is not in the denominator.

\begin{aligned}& \cfrac{1}{x}+\cfrac{3}{2 x}=6 \\\\ & \cfrac{1}{\cfrac{5}{12}}+\cfrac{3}{2\left(\cfrac{5}{12}\right)}=6 \\\\ & \cfrac{12}{5}+\cfrac{3}{\cfrac{10}{12}}=6 \\\\ & \cfrac{12}{5}+\cfrac{12}{10}\left(\cfrac{3}{1}\right)=6 \\\\ & \cfrac{12}{5}+\cfrac{36}{10}=6 \\\\ & \cfrac{24}{10}+\cfrac{36}{10}=6 \\\\ & \cfrac{60}{10}=6 \; ✅\end{aligned}

What are rational functions?

Common Core State Standards

How does this relate to high school math?

High School Algebra: Arithmetic with Polynomials and Rational Expressions (HSA-APR.D.6)
Rewrite simple rational expressions in different forms; write $\cfrac{a(x)}{b(x)}$ in the form $q(x)+ \cfrac{r(x)}{b(x)},$ where $a(x), b(x), q(x),$ and $r(x)$ are polynomials with the degree of $r(x)$ less than the degree of $b(x),$ using inspection, long division, or, for the more complicated examples, a computer algebra system.

High School Algebra: Arithmetic with Polynomials and Rational Expressions (HSA-APR.D.7)
Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions.

How to simplify and apply operations to rational functions

If you want more detailed steps and practice for simplifying rational expressions and applying the operations to rational expressions, check out the links highlighted above or take a look at the examples below.

Simplify and apply operations of rational functions examples

Example 1: simplify rational expressions without factoring

Simplify the expression: $\cfrac{28 x^2 y}{7 x y^2}$ .

Identify the greatest common factor of the numerator and the denominator.

The numerator, $28 x^2 y$ and the denominator, $7 x y^2$ have a greatest common factor of, $7 x y.$

Looking at the coefficients of $28$ and $7,$ the greatest common factor between those numbers is $7.$

Looking at the variable $x,$ the greatest common factor between $x^2$ and $x$ is $x.$

Lastly, looking at the variable y, the greatest common factor between $y$ and $y^2$ is $y.$

So, the greatest common factor between $28 x^2 y$ and $7 x y^2$ is $7 x y.$

2Simplify the expression by the greatest common factor.

3Write the simplified expression.

\cfrac{28 x^2 y}{7 x y^2}=\cfrac{4 x}{y}

Example 2: subtracting rational expressions by factoring

Subtract the rational expressions, $\cfrac{8}{(x-2)}-\cfrac{9}{\left(x^2-4\right)}$ .

If possible, factor the denominators before finding the least common denominator.

$(x-2) \rightarrow$ Cannot be factored

$\left(x^2-4\right) \rightarrow$ Can be factored because it the difference of two perfect squares.

Applying the strategy of factoring the difference between two perfect squares, $\left(x^2-4\right)$ factors to be $(x+2)(x-2).$

Find the least common denominator.

After factoring,

$\cfrac{8}{(x-2)}-\cfrac{9}{\left(x^2-4\right)}=\cfrac{8}{(x-2)}-\cfrac{9}{(x+2)(x-2)}$

Looking at the factored expression, the denominators are:

\begin{aligned}& (x-2) \\\\ & (x+2)(x-2) \end{aligned}

So, the least common denominator is $(x+2)(x-2),$ take one of each factor to get the common denominator.

Adjust the rational expressions to have the common denominator.

\begin{aligned}&\cfrac{(x+2)}{(x+2)} \times \cfrac{8}{(x-2)}-\cfrac{9}{(x+2)(x-2)} \\\\ &\begin{aligned}& \cfrac{8(x+2)}{(x+2)(x-2)}-\cfrac{9}{(x+2)(x-2)} \\\\ & \cfrac{8 x+16}{(x+2)(x-2)}-\cfrac{9}{(x+2)(x-2)} \end{aligned} \end{aligned}

Subtract and simplify the rational expression.

\cfrac{8 x+16}{(x+2)(x-2)}-\cfrac{9}{(x+2)(x-2)}=\cfrac{8 x+7}{(x+2)(x-2)}

$\cfrac{8 x+7}{(x+2)(x-2)} \rightarrow$ The difference cannot be simplified further, but the denominator can be multiplied back out.

$\cfrac{8 x+7}{(x+2)(x-2)}=\cfrac{8 x+7}{\left(x^2-4\right)}$

Example 3: multiplying without factoring

Multiply the rational expressions: $\cfrac{8 a}{14 b^2} \times \cfrac{7 b^3}{20 a^2}$ .

Factor the expression if possible.

The expression cannot be factored.

Simplify the expression before multiplying.

\cfrac{8 a}{14 b^2} \times \cfrac{7 b^3}{20 a^2}

In this case, $7$ and $14$ can simplify by a common factor of $7; b^2$ and $b^3$ simplify by a common factor of $b^2.$

Also, $8$ and $20$ can simplify by a factor of $4; a$ and $a^2$ can simplify by a factor of $a.$

Multiply the expressions and simplify the product if necessary.

$\cfrac{2}{2} \times \cfrac{b}{5 a}=\cfrac{2 b}{10 a} \rightarrow$ The product can be simplified further because $2$ and $10$ have a common factor of $2.$

So, $\cfrac{2 b}{10 a}=\cfrac{b}{5 a}$

Example 4: divide rational expressions with factoring

Divide the rational functions, $\cfrac{5}{\left(x^2+5 x+6\right)} \div \cfrac{4}{\left(x^2+6 x+9\right)}$ .

Factor the expression if possible.

The denominators of both rational expressions can be factored.

\begin{aligned}& \cfrac{5}{\left(x^2+5 x+6\right)} \div \cfrac{4}{\left(x^2+6 x+9\right)} \\\\ & \cfrac{5}{(x+2)(x+3)} \div \cfrac{4}{(x+3)(x+3)} \end{aligned}

With division, take the reciprocal of the second expression and change the operation to multiplication.

The reciprocal of $\cfrac{4}{(x+3)(x+3)}$ is $\cfrac{(x+3)(x+3)}{4}$

The new expression is:

$\cfrac{5}{(x+2)(x+3)} \times \cfrac{(x+3)(x+3)}{4}$

Simplify before multiplying.

Look for common factors to cancel.

The only common factor was $x+3$ .

Multiply the expressions and simplify the answer if possible.

\cfrac{5}{(x+2)} \times \cfrac{(x+3)}{4}=\cfrac{5(x+3)}{4(x+2)}=\cfrac{5 x+3}{4 x+8}

You can leave the answer as $\cfrac{5(x+3)}{4(x+2)}$ or apply the distributive property, $\cfrac{5 x+3}{4 x+8}$

How to solve rational equations

In order to solve rational equations:

Find the common denominator of all the rational expressions in the equation.
Multiply the entire equation by the common denominator.
Solve the equation.
Check the solution

Solve rational equation examples

Example 5: equation with x in the denominator

Solve the equation:

\cfrac{5}{2 x}-\cfrac{14}{x}=3

Find the common denominator of all the rational expressions in the equation.

The denominators are $2x , x$ and $1$ (whole numbers have a denominator of $1$ ).

The common denominator which is also the least common multiple of all expressions, is $2x.$

Multiply the entire equation by the common denominator.

$2 x\left(\cfrac{5}{2x}-\cfrac{14}{x}=3\right)$

\begin{aligned}& 2 x \times \cfrac{5}{2 x}=5 \\\\ & 2 x \times \cfrac{14}{x}=28 \\\\ & 2 x \times 3=6 x \end{aligned}

Solve the equation.

After multiplying by $2x,$ the new equation is:

\begin{aligned}& 5-28=6 x \\\\ & - \, 23=6 x \\\\ & \cfrac{- \, 23}{6}=\cfrac{6 x}{6} \\\\ & \cfrac{- \, 23}{6}=- \, 3 \cfrac{5}{6}=x \end{aligned}

Check the solution.

Substitute $\cfrac{- \, 23}{6}$ for $x$ into the original equation.

\begin{aligned} &\begin{aligned}& \cfrac{5}{2 x}-\cfrac{14}{x}=3 \\\\ & \cfrac{5}{2\left(\cfrac{- \, 23}{6}\right)}-\cfrac{14}{\cfrac{- \, 23}{6}}=3\end{aligned} \\\\ &\begin{aligned}& \cfrac{5}{\cfrac{- \, 23}{3}}-\cfrac{14}{\cfrac{- \, 23}{6}}=3 \\\\ & 5 \times\left(\cfrac{3}{- \, 23}\right)-14 \times\left(\cfrac{6}{- \, 23}\right)=3 \\\\ & \cfrac{15}{- \, 23}-\cfrac{84}{- \, 23}=3 \\\\ & - \, \cfrac{15}{23}+\cfrac{84}{23}=3 \\\\ & \cfrac{69}{23}=3 \\\\ & 3=3 \; ✅ \end{aligned} \end{aligned}

Example 6: distributing binomials

\cfrac{x+1}{x+2}+\cfrac{3}{x-4}=1

Find the common denominator of all the rational expressions in the equation.

The denominators are: $(x+2), (x-4),$ and $1$

The common denominator is $(x+2)(x-4)$ .

Multiply the entire equation by the common denominator.

$(x+2)(x-4)\left[\cfrac{x+1}{x+2}+\cfrac{3}{x-4}=1\right]$

\begin{aligned}& (x+2)(x-4)\left(\cfrac{x+1}{x+2}\right)=(x-4)(x+1)=x^2-3 x-4 \\\\ & (x+2)(x-4)\left(\cfrac{3}{x-4}\right)=3(x+2)=3 x+6 \end{aligned}

$(x+2)(x-4)(1)=(x+2)(x-4)=x^2-2 x-8$

Solve the equation.

After multiplying by the common denominator, $(x+2)(x-4)$ the equation is:

\begin{aligned}& x^2-3 x-4+3 x+6=x^2-2 x-8 \\\\ & x^2-x^2-3 x-4+3 x+6=x^2-x^2-2 x-8 \\\\ & - \, 3 x-4+3 x+6=- \, 2 x-8 \\\\ & \quad 2=- \, 2 x-8 \\\\ & 2+8=- \, 2 x-8+8 \\\\ & 10=- \, 2 x \\\\ & \cfrac{10}{- \, 2}=\cfrac{- \, 2 x}{- \, 2} \\\\ & - \, 5=x \end{aligned}

Check the solution

Substitute $5$ for $x$ into the original equation.

\begin{aligned}& \cfrac{x+1}{x+2}+\cfrac{3}{x-4}=1 \\\\ & \cfrac{- \, 5+1}{- \, 5+2}+\cfrac{3}{- \, 5-4}=1 \\\\ & \cfrac{- \, 4}{- \, 3}+\cfrac{3}{- \, 9}=1 \\\\ & \cfrac{4}{3}-\cfrac{3}{9}=1 \\\\ & \cfrac{4}{3}-\cfrac{1}{3}=1 \\\\ & \cfrac{3}{3}=1 \\\\ & 1=1 \; ✅ \end{aligned}

Teaching tips for rational functions

Make connections between operations with fractions to operations with rational functions so that students can develop understanding.

Use active learning activities such as scavenger hunts or game playing to have students review skills instead of giving them a worksheet.

Encourage students who are struggling to use digital platforms such as Khan Academy so they can view tutorial videos.

Easy mistakes to make

Multiplying the numerator and the denominator by the common denominator
When solving rational equations, you have to multiply the entire equation by the common denominator. For example,

$\cfrac{3}{2x}-\cfrac{1}{5x}=11$

The common denominator is $10x,$ so multiply the entire equation by $10x.$

$10 x\left(\cfrac{3}{2 x}-\cfrac{1}{5 x}=11\right),$ when performing this multiplication,

$10 x \times \cfrac{3}{2 x}$ ≠ $\cfrac{30 x}{20 x}$ it should be $10 x \times \cfrac{3}{2 x}=15$

So then, $10 x \times \cfrac{1}{5 x}=2$ and $10 x \times 11=110 x$

Not multiplying all terms by the denominator
When multiplying the equation by the common denominator, be sure to multiply every term.

$\begin{aligned}& 3-\cfrac{5}{x-1}=1 \\\\ & (x-1)\left[3-\cfrac{5}{x-1}=1\right] \end{aligned}$

$3(x-1)-5=1 \rightarrow$ The last term was not multiplied by $(x-1)$

So, it should be, $3(x-1)-5=x-1$ .

Adding rational expressions incorrectly
For example, thinking that you should add the numerators and the denominators,

$\cfrac{8}{4 x}+\cfrac{7}{x}$ ≠ $\cfrac{15}{5 x}$

In order to add them, you must find a common denominator just like when you add fractions. In this case, the common denominator is $4x.$

$\begin{aligned}& \cfrac{8}{4 x}+\cfrac{4}{4} \times \cfrac{7}{x}= \\\\ & \cfrac{8}{4 x}+\cfrac{28}{4 x}=\cfrac{36}{4 x} \end{aligned}$
The answer can be simplified further to be, $\cfrac{36}{4 x}=\cfrac{9}{x}$ .

Practice rational functions questions

(x+7)

\cfrac{1}{(x+7)}

\cfrac{1}{(x+2)}

\cfrac{1}{(x+2)(x+7)}

Factor the numerator and denominator before simplifying.

The numerator cannot be factored in this case, but the denominator can be factored.

\cfrac{(x+7)}{\left(x^2+9 x+14\right)}=\cfrac{x+7}{(x+7)(x+2)}

After factoring, cancel the terms in the numerator that are the same as the terms in the denominator.

Rational Functions 11 US

The simplified expression is: $\cfrac{1}{x+2}$ Remember that there is a $1$ in the numerator because $\cfrac{x+7}{x+7}=\cfrac{1}{1}$

\cfrac{5 x+12}{(x+2)(x-1)}

\cfrac{15 x+12}{(x+2)(x-1)}

\cfrac{3 x+24}{(x+2)(x-1)}

\cfrac{15}{(x+2)(x-1)}

In order to add rational functions, you have to find the common denominator.

The common denominator of $\cfrac{6}{(x+2)}+\cfrac{9}{(x-1)}$ is $(x+2)(x-1)$

Multiply the numerator and denominator by the common denominator.

\begin{aligned}&\cfrac{(x+2)(x-1)}{(x+2)(x-1)}\left(\cfrac{6}{x+2}\right)=\cfrac{6(x-1)}{(x+2)(x-1)} \\\\ & \cfrac{(x+2)(x-1)}{(x+2)(x-1)}\left(\cfrac{9}{x-1}\right)=\cfrac{9(x+2)}{(x+2)(x-1)} \end{aligned}

\cfrac{6(x-1)}{(x+2)(x-1)}+\cfrac{9(x+2)}{(x+2)(x-1)}=\cfrac{6(x-1)+9(x+2)}{(x+2)(x-1)}=\cfrac{6 x-6+9 x+18}{(x+2)(x-1)}=\cfrac{15 x+12}{(x+2)(x-1)}

\cfrac{30 x y}{y}

\cfrac{150 x y}{5 y^2}

\cfrac{30 x^3 y}{x}

30 x^2 y^2

You can either simplify first and then multiply or multiply first and then simplify. In this case, let’s simplify first:

Rational Functions 12 US

\cfrac{6 y^2}{1} \times \cfrac{5 x^2}{1}=\cfrac{30 x^2 y^2}{1}=30 x^2 y^2

\cfrac{x+8}{4}

\cfrac{4}{x+8}

\cfrac{100}{(x-8)(x-8)(x+8)}

\cfrac{x-8}{4}

When dividing rational expressions, remember to take the reciprocal of the second expression and multiply it to the first.

\cfrac{5}{(x-8)} \div \cfrac{20}{\left(x^2-64\right)}=\cfrac{5}{(x-8)} \times \cfrac{\left(x^2-64\right)}{20}

Before multiplying, factor and simplify the expressions by canceling any terms that match in the numerator with terms in the denominator.

\cfrac{5}{(x-8)} \times \cfrac{(x-8)(x+8)}{20}

Rational Functions 13 US

Now, multiply: $\cfrac{1}{1} \times \cfrac{(x+8)}{4}=\cfrac{x+8}{4}$

\cfrac{9}{16}

– \, \cfrac{9}{16}

\cfrac{16}{9}

– \, \cfrac{16}{9}

Multiply the entire equation by the common denominator.

In this case, the common denominator is $4x.$

\begin{aligned}& 4 x\left(\cfrac{4}{x}+\cfrac{2}{4 x}=- \, 8\right) \\\\ & 16+2=- \, 32 x \\\\ & 18=- \, 32 x \\\\ & \cfrac{18}{- \, 32}=\cfrac{- \, 32 x}{- \, 32} \\\\ & – \, \cfrac{18}{32}=x \end{aligned}

Checking the solution, it works in the original equation.

\begin{aligned}& \cfrac{4}{\cfrac{- \, 18}{32}}+\cfrac{2}{4\left(\cfrac{- \, 18}{32}\right)}=- \, 8 \\\\ & 4 \times- \, \cfrac{32}{18}+\cfrac{2}{\cfrac{- \, 18}{8}}=- \, 8 \\\\ & – \, \cfrac{64}{9}+\left(- \, \cfrac{8}{9}\right)=- \, 8 \\\\ & – \, \cfrac{72}{9}=- \, 8 \\\\ & – \, 8=- \, 8 \; ✅ \end{aligned}

$x=2.5$ or $x=5$

x=- \, 1

x=\sqrt{11}

x=0

Find the common denominator of the equation and multiply the entire equation by the common denominator. In this case, the common denominator is $5x.$

\begin{aligned}& 5 x\left(\cfrac{5}{x}+\cfrac{2 x}{5}=3\right) \\\\ & 5 x \times \cfrac{5}{x}=25 \\\\ & 5 x \times \cfrac{2 x}{5}=2 x^2 \\\\ & 5 x \times 3=15 x \end{aligned}

After multiplying, the new equation is quadratic.

25+2 x^2=15 x

Solve the quadratic by first setting the quadratic equation equal to $0$ and then using factoring or the quadratic formula to solve.

If you need extra help with solving quadratic equations, check out these links.

How to factor quadratic equations

Quadratic formula

Solving quadratic equations

\begin{aligned} &\begin{aligned}& 25+2 x^2-15 x=15 x-15 x \\\\ & 25+2 x^2-15 x=0 \\\\ & 2 x^2-15 x+25=0 \\\\ & (2 x-5)(x-5)=0 \\\\ & (2 x-5)=0 \hspace{1.85cm} (x-5)=0 \end{aligned} \\\\ &\begin{aligned}& 2 x-5+5=0+5 \hspace{1cm}  x-5+5=0+5\\\\ & 2 x=5 \hspace{2.7cm} x=5 \\\\ & \cfrac{2 x}{2}=\cfrac{5}{2} \\\\ & x=\cfrac{5}{2}=2.5\end{aligned} \end{aligned}

Checking both values, they work in the original equation.

\begin{aligned}& \cfrac{5}{x}+\cfrac{2 x}{5}=3 \\\\ & \cfrac{5}{2.5}+\cfrac{2 \times 2.5}{5}=3 \\\\ & 2+1=3 \; ✅ \\\\ & \cfrac{5}{x}+\cfrac{2 x}{5}=3 \\\\ & \cfrac{5}{5}+\cfrac{2 \times 5}{5}=3 \\\\ & 1+2=3 \; ✅ \end{aligned}

Rational functions FAQs

What are the asymptotes of rational functions?

The asymptotes of rational functions are horizontal lines, vertical lines, or slanted lines that the graph of a rational function approaches but does not cross.

Typically, when sketching the graph of the rational function by hand, the horizontal and vertical asymptotes, as well as the slant asymptotes are noted with a dotted line.

How do you find the horizontal and vertical asymptotes of rational functions?

Thinking about the function values that exist in the domain of a rational function and those that do not, helps you find the horizontal and vertical asymptotes.

The values for $x$ that make the denominator equal to $0$ are the possible values for the vertical asymptote.

The horizontal asymptote is defined by the ratio of the leading coefficients of the numerator and denominator.

Do rational functions have inverse functions?

Yes, you can find the inverse functions of rational functions. This concept will be explored in a precalculus course.

How do you find the $\textbf{x}$ -intercept and the $\textbf{y}$ -intercept of rational functions?

You find the $x$ -intercepts and the $y$ -intercepts of rational functions applying the same strategy as you would with any function.

For the $x$ -intercept set the function equal to $0$ and find the $x$ values.

For the $y$ -intercept substitute $0$ in for the $x$ values and find $y.$ The $x$ and $y$ intercepts are real numbers.

What is the reciprocal function?

The reciprocal function, $f(x)=\cfrac{1}{x}$ is defined to be the reciprocal function, which is also a rational function.

Is the equation of a circle a rational function?

No, the equation of a circle is not a rational function or even a function at all, it’s considered to be a conic section.

What are matrices?

Matrices are a rectangular array of numbers and can be used to solve systems of linear equations.

The next lessons are

Vectors

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