Problem 59
Question
Find the domain of each function. $$ H(r)=\frac{4}{r^{2}+11 r+24} $$
Step-by-Step Solution
Verified Answer
The domain of the function \(H(r)=\frac{4}{r^{2}+11 r+24}\) is all real numbers except -8 and -3.
1Step 1: Set the denominator equal to zero
In order to find the r-values that are not allowed, we need to set the denominator equal to zero and solve. That is done as follows: \(r^{2}+11 r+24 = 0\).
2Step 2: Factor the equation
After setting the denominator to zero, we need to solve the quadratic equation. It can be factored as follows: \((r+8)(r+3) = 0\).
3Step 3: Find the roots
By setting each factor equal to zero, we can solve for r. This yields the roots: \(r+8 = 0 => r=-8\) and \(r+3 = 0 => r=-3\).
4Step 4: Determine the domain
The r-values -8 and -3 make the denominator of our function equal to zero and therefore we must exclude these from our domain. This means that the domain of the function is all real numbers except -8 and -3.
Key Concepts
Algebraic FunctionsQuadratic EquationsFactoring Polynomials
Algebraic Functions
In mathematics, an algebraic function is a type of function that can be constructed using algebraic operations such as addition, subtraction, multiplication, division, and root extraction. These functions are generally defined by a polynomial equation or a ratio of two polynomials, known as a rational function.
When dealing with algebraic functions, an important aspect is to determine their domain, which consists of all the possible input values (usually represented by x or r) that can be used within the function without causing mathematical issues such as division by zero.
To find the domain of a function like H(r) in our exercise, it's essential to ensure the denominator never equals zero. Since the denominator is a polynomial, these problematic values, also called excluded values, can be determined using skills in solving quadratic equations and factoring polynomials.
When dealing with algebraic functions, an important aspect is to determine their domain, which consists of all the possible input values (usually represented by x or r) that can be used within the function without causing mathematical issues such as division by zero.
To find the domain of a function like H(r) in our exercise, it's essential to ensure the denominator never equals zero. Since the denominator is a polynomial, these problematic values, also called excluded values, can be determined using skills in solving quadratic equations and factoring polynomials.
Quadratic Equations
A quadratic equation is a second-degree polynomial equation in the form of \(ax^2 + bx + c = 0\), where a, b, and c are constants, and a is not equal to zero. These equations are fundamental in algebra and can have two real solutions, one real solution, or no real solutions when considering real numbers only.
To solve quadratic equations, one can use methods such as factoring, completing the square, or the quadratic formula. In the context of finding the domain of a function, like in our textbook exercise, we factor the quadratic equation in the denominator to discover values that are not allowed in the domain. Solving these equations provides crucial insights into the behavior of algebraic functions.
To solve quadratic equations, one can use methods such as factoring, completing the square, or the quadratic formula. In the context of finding the domain of a function, like in our textbook exercise, we factor the quadratic equation in the denominator to discover values that are not allowed in the domain. Solving these equations provides crucial insights into the behavior of algebraic functions.
Factoring Polynomials
Factoring polynomials is a critical skill in algebra that allows for the simplification of equations and easier computation of zeros or roots. To factor a polynomial is to break it down into the product of its simpler factors. This can often simplify complex algebraic expressions and equations.
The process often involves finding two numbers that both add up to the coefficient of the linear term (b in a quadratic equation) and multiply to the constant term (c). In step 2 from our provided solutions, the quadratic equation \(r^2 + 11r + 24 = 0\) was factored into \(r + 8)(r + 3) = 0\). These factors reveal that the values r = -8 and r = -3 are the roots of the polynomial, meaning they make the original quadratic expression zero, and thus, can't be part of the function's domain.
The process often involves finding two numbers that both add up to the coefficient of the linear term (b in a quadratic equation) and multiply to the constant term (c). In step 2 from our provided solutions, the quadratic equation \(r^2 + 11r + 24 = 0\) was factored into \(r + 8)(r + 3) = 0\). These factors reveal that the values r = -8 and r = -3 are the roots of the polynomial, meaning they make the original quadratic expression zero, and thus, can't be part of the function's domain.
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