Problem 96 - Beginning and Intermediate Algebra

Question

If \(h(a)=-a^{2}-3 a+10,\) find a) \(h(5)\) b) \(h(-4)\)

Step-by-Step Solution

Answer

a) \(h(5) = -30\) b) \(h(-4) = 6\)

Step 1: Substitute the value of 'a' in the function h(a)

To find the value of h(5), we will substitute 'a' with 5 in the given function: h(a) = -a^2 - 3a + 10.

Step 2: Solve for h(5)

Now that we have substituted the value of 'a', we can calculate h(5): h(5) = -(5)^2 - (3)(5) + 10 h(5) = -25 - 15 + 10 h(5) = -40 + 10 h(5) = -30 b) Finding h(-4)

Step 1: Substitute the value of 'a' in the function h(a)

To find the value of h(-4), we will substitute 'a' with -4 in the given function: h(a) = -a^2 - 3a + 10.

Step 2: Solve for h(-4)

Now that we have substituted the value of 'a', we can calculate h(-4): h(-4) = -(-4)^2 - (3)(-4) + 10 h(-4) = -16 + 12 + 10 h(-4) = -4 + 10 h(-4) = 6 The final answers are: a) h(5) = -30 b) h(-4) = 6

Key Concepts

Algebraic Functions

Algebraic functions are functions that can be expressed using basic algebraic operations such as addition, subtraction, multiplication, division, and taking roots. These functions are represented by a formula that involves variables and coefficients. In the given problem, the function is defined as \(h(a) = -a^2 - 3a + 10\), which clearly shows an algebraic structure.
Algebraic functions are fundamental in mathematics because they provide a way to represent real-world relationships and various phenomena. When analyzing such functions, you often need to substitute specific values for the variables to evaluate or solve equations, a common technique in algebra. Understanding how to manipulate and substitute values into algebraic expressions is crucial for solving problems like the one provided. The operations involved in algebraic functions follow the order of operations, often remembered by PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction). Remembering this rule will help you avoid errors when evaluating these functions.

Substitution Method

The substitution method is a straightforward and essential technique when working with functions. It involves replacing a variable in a function with a given number to find the function's value at that point. For example, to find \(h(5)\), we substitute \(a\) with 5 in the function. This transforms \(h(a) = -a^2 - 3a + 10\) into \(h(5) = -(5)^2 - 3(5) + 10\).

• Substitute the given value into the function equation.
• Simplify the expression using the order of operations.
• Calculate the final value to find the function's output at the specified input.

The substitution method is widely used in different areas such as evaluating limits, solving equations, and checking solutions. By practicing this process, students can become more adept at handling complex expressions and equations. Using substitutions can also help simplify expressions, making them easier to understand and work with.

Polynomial Functions

Polynomial functions are a specific type of algebraic function which consist entirely of terms in the form \(ax^n\), where \(a\) is a coefficient and \(n\) is a nonnegative integer. In our exercise, the function \(h(a) = -a^2 - 3a + 10\) is a polynomial of degree 2 because the highest power of \(a\) is 2.
Key characteristics of polynomial functions include:

• Degree: Indicates the highest exponent of the variable in the polynomial. Here, \(a^2\) makes it a quadratic function.
• Coefficients: The numbers multiplying the variables, i.e., -1 for \(a^2\) and -3 for \(a\) in our example.
• Constant terms: The number at the end of the polynomial without any variable, which is 10 here.

Polynomial functions are used extensively in physics, engineering, economics, and many other fields to model behaviors and draw conclusions from data. Evaluating polynomial functions often involves substituting specific values to calculate outputs, as demonstrated in solving for \(h(5)\) and \(h(-4)\). Understanding the structure of polynomials helps in graphing functions, predicting trends, and solving practical problems.