Problem 76
Question
Find each function value. \(h(x)=2 x^{2}-4, h(0)\)
Step-by-Step Solution
Verified Answer
The value of the function \(h(0)\) is \(-4\).
1Step 1: Understand the Problem
We need to find the value of the function \(h(x) = 2x^2 - 4\) when \(x\) is equal to 0.
2Step 2: Substitute in the Function
Substitute \(x = 0\) into the function. The function is \(h(x) = 2x^2 - 4\). So, replace \(x\) with 0: \(h(0) = 2(0)^2 - 4\).
3Step 3: Simplify the Expression
Calculate \((0)^2\), which is 0, and substitute it back into the equation: \(h(0) = 2 \times 0 - 4\).
4Step 4: Solve the Expression
Multiply 2 by 0, which gives 0: \(h(0) = 0 - 4\). Continue by simplifying the subtraction: \(h(0) = -4\).
5Step 5: Conclusion
Thus, the value of the function \(h(0)\) is \(-4\).
Key Concepts
SubstitutionAlgebraic ExpressionsQuadratic Functions
Substitution
In mathematics, substitution is a fundamental concept often used to calculate values, solve equations, or evaluate expressions by replacing variables with numbers or other expressions. When we talk about function evaluation, substitution plays a crucial role.
To evaluate the function \(h(x) = 2x^2 - 4\) at \(x = 0\), we substitute \(0\) for \(x\) in the expression. This means every place where \(x\) appears in the equation, it is replaced with \(0\).
For instance:
To evaluate the function \(h(x) = 2x^2 - 4\) at \(x = 0\), we substitute \(0\) for \(x\) in the expression. This means every place where \(x\) appears in the equation, it is replaced with \(0\).
For instance:
- Start with the function: \(h(x) = 2x^2 - 4\).
- Substitute \(x = 0\): \(h(0) = 2(0)^2 - 4\).
Algebraic Expressions
Algebraic expressions are combinations of numbers, variables, and operators such as addition, subtraction, multiplication, and division. These are the building blocks of algebra and are used to represent relationships and calculations concisely.
In the given exercise, the function \(h(x) = 2x^2 - 4\) is an algebraic expression. It includes:
In the given exercise, the function \(h(x) = 2x^2 - 4\) is an algebraic expression. It includes:
- Terms: \(2x^2\) and \(-4\)
- Coefficient: the number \(2\) which is multiplied by the variable part \(x^2\)
- Constant: \(-4\), a term that remains unchanged since it lacks a variable part
Quadratic Functions
Quadratic functions are polynomial functions characterized by the highest power of the variable being two. A general form of a quadratic function is \(ax^2 + bx + c\).
In our exercise, the function \(h(x) = 2x^2 - 4\) is a classic example of a quadratic function. This specific function is expressed with:
In our exercise, the function \(h(x) = 2x^2 - 4\) is a classic example of a quadratic function. This specific function is expressed with:
- A leading coefficient \(a\) of \(2\), attached to \(x^2\)
- Zero for the linear term \(b\)
- A constant term \(c\) of \(-4\)
Other exercises in this chapter
Problem 75
If \(y\) varies directly as \(x\) and \(y=5\) when \(x=2,\) find \(y\) when \(x=6\)
View solution Problem 75
Simplify. $$ \sqrt{26} \cdot \sqrt{39} \cdot \sqrt{14} $$
View solution Problem 76
Factor completely. If the polynomial is not factorable, write prime. $$ 13 x y z+3 x^{2} z+4 k $$
View solution Problem 76
Simplify each expression. \(\frac{39 a^{3} b^{4}}{13 a^{4} b^{3}}\)
View solution