Given The Function F ( X ) = X 2 − 2 X − 24 X 2 + 10 X + 24 F(x)=\frac{x^2-2x-24}{x^2+10x+24} F ( X ) = X 2 + 10 X + 24 X 2 − 2 X − 24 ​ , Determine The Nature Of F F F At Each Of The Following Values Of X X X : Whether It Has A Zero, A Vertical Asymptote, Or A Removable

Introduction

In mathematics, rational functions are a type of function that can be expressed as the ratio of two polynomials. These functions can exhibit various behaviors, including zeros, vertical asymptotes, and removable discontinuities. In this article, we will analyze the nature of the rational function $f(x)=\frac{x^2-2x-24}{x^2+10x+24}$ at specific values of $x$.

Understanding the Function

The given function is a rational function, which means it is the ratio of two polynomials. The numerator is $x^2-2x-24$, and the denominator is $x^2+10x+24$. To analyze the nature of this function, we need to examine the behavior of the numerator and denominator separately.

Factoring the Numerator and Denominator

Let's start by factoring the numerator and denominator.

Factoring the Numerator

The numerator can be factored as follows:

$x^2-2x-24 = (x-8)(x+3)$

Factoring the Denominator

The denominator can be factored as follows:

$x^2+10x+24 = (x+2)(x+12)$

Analyzing the Zeros

A zero of a function is a value of $x$ that makes the function equal to zero. To find the zeros of the function, we need to set the numerator equal to zero and solve for $x$.

Finding the Zeros

Setting the numerator equal to zero, we get:

$(x-8)(x+3) = 0$

This equation has two solutions: $x=8$ and $x=-3$. Therefore, the function has zeros at $x=8$ and $x=-3$.

Analyzing the Vertical Asymptotes

A vertical asymptote of a function is a value of $x$ that makes the function undefined. To find the vertical asymptotes, we need to set the denominator equal to zero and solve for $x$.

Finding the Vertical Asymptotes

Setting the denominator equal to zero, we get:

$(x+2)(x+12) = 0$

This equation has two solutions: $x=-2$ and $x=-12$. Therefore, the function has vertical asymptotes at $x=-2$ and $x=-12$.

Analyzing the Removable Discontinuities

A removable discontinuity of a function is a value of $x$ that makes the function undefined, but the limit of the function as $x$ approaches that value exists. To find the removable discontinuities, we need to examine the behavior of the function as $x$ approaches the values that make the denominator equal to zero.

Finding the Removable Discontinuities

As $x$ approaches $-2$, the function approaches:

$\lim_{x\to-2} \frac{(x-8)(x+3)}{(x+2)(x+12)} = \frac{(-2-8)(-2+3)}{(-2+2)(-2+12)} = \frac{-10}{0}$

This limit does not exist, so there is no removable discontinuity at $x=-2$.

As $x$ approaches $-12$, the function approaches:

$\lim_{x\to-12} \frac{(x-8)(x+3)}{(x+2)(x+12)} = \frac{(-12-8)(-12+3)}{(-12+2)(-12+12)} = \frac{-120}{0}$

This limit does not exist, so there is no removable discontinuity at $x=-12$.

However, we can simplify the function by canceling out the common factors in the numerator and denominator.

Simplifying the Function

The function can be simplified as follows:

$f(x) = \frac{(x-8)(x+3)}{(x+2)(x+12)} = \frac{x-8}{x+12}$

This simplified function has a removable discontinuity at $x=-12$, since the limit of the function as $x$ approaches $-12$ exists.

Conclusion

In conclusion, the rational function $f(x)=\frac{x^2-2x-24}{x^2+10x+24}$ has zeros at $x=8$ and $x=-3$, vertical asymptotes at $x=-2$ and $x=-12$, and a removable discontinuity at $x=-12$. The function can be simplified by canceling out the common factors in the numerator and denominator.

References

• [1] "Rational Functions" by Math Open Reference
• [2] "Analyzing Rational Functions" by Khan Academy

Further Reading

• "Rational Functions" by Wolfram MathWorld
• "Analyzing Rational Functions" by MIT OpenCourseWare
  Q&A: Analyzing Rational Functions =====================================

Introduction

In our previous article, we analyzed the nature of the rational function $f(x)=\frac{x^2-2x-24}{x^2+10x+24}$. We found that the function has zeros at $x=8$ and $x=-3$, vertical asymptotes at $x=-2$ and $x=-12$, and a removable discontinuity at $x=-12$. In this article, we will answer some frequently asked questions about analyzing rational functions.

Q: What is a rational function?

A rational function is a type of function that can be expressed as the ratio of two polynomials. It is a function that has a numerator and a denominator, and the denominator is not equal to zero.

Q: How do I find the zeros of a rational function?

To find the zeros of a rational function, you need to set the numerator equal to zero and solve for $x$. This will give you the values of $x$ that make the function equal to zero.

Q: How do I find the vertical asymptotes of a rational function?

To find the vertical asymptotes of a rational function, you need to set the denominator equal to zero and solve for $x$. This will give you the values of $x$ that make the function undefined.

Q: What is a removable discontinuity?

A removable discontinuity is a value of $x$ that makes the function undefined, but the limit of the function as $x$ approaches that value exists. This means that the function can be simplified by canceling out the common factors in the numerator and denominator.

Q: How do I simplify a rational function?

To simplify a rational function, you need to cancel out the common factors in the numerator and denominator. This will give you a simpler function that has the same zeros and vertical asymptotes as the original function.

Q: What are some common mistakes to avoid when analyzing rational functions?

Some common mistakes to avoid when analyzing rational functions include:

• Not factoring the numerator and denominator correctly
• Not canceling out common factors in the numerator and denominator
• Not checking for removable discontinuities
• Not simplifying the function correctly

Q: How do I use technology to analyze rational functions?

There are many online tools and software programs that can help you analyze rational functions. Some popular options include:

• Graphing calculators
• Online graphing tools
• Computer algebra systems (CAS)
• Mathematical software programs

Q: What are some real-world applications of rational functions?

Rational functions have many real-world applications, including:

• Physics: Rational functions are used to model the motion of objects and the behavior of physical systems.
• Engineering: Rational functions are used to design and analyze electrical circuits and mechanical systems.
• Economics: Rational functions are used to model economic systems and make predictions about future economic trends.
• Computer Science: Rational functions are used in computer graphics and game development.

Conclusion

In conclusion, analyzing rational functions is an important topic in mathematics and has many real-world applications. By understanding the nature of rational functions, you can solve problems and make predictions about complex systems. We hope this Q&A article has been helpful in answering some of your questions about analyzing rational functions.

References

• [1] "Rational Functions" by Math Open Reference
• [2] "Analyzing Rational Functions" by Khan Academy
• [3] "Rational Functions" by Wolfram MathWorld
• [4] "Analyzing Rational Functions" by MIT OpenCourseWare

Further Reading

• "Rational Functions" by Cambridge University Press
• "Analyzing Rational Functions" by Springer-Verlag
• "Rational Functions" by Oxford University Press