Problem 20 - Algebra and Trigonometry

Question

Evaluate the function at the indicated values. $$ \begin{array}{l}{f(x)=x^{2}+2 x} \\ {f(0), f(3), f(-3), f(a), f(-x), f\left(\frac{1}{a}\right)}\end{array} $$

Step-by-Step Solution

Answer

f(0) = 0, f(3) = 15, f(-3) = 3, f(a) = a^2 + 2a, f(-x) = x^2 - 2x, f(1/a) = 1/a^2 + 2/a.

Step 1: Evaluate f(0)

Substitute 0 into the function: \[ f(0) = 0^2 + 2 \times 0 = 0 \] Therefore, \( f(0) = 0 \).

Step 2: Evaluate f(3)

Substitute 3 into the function: \[ f(3) = 3^2 + 2 \times 3 = 9 + 6 = 15 \] Therefore, \( f(3) = 15 \).

Step 3: Evaluate f(-3)

Substitute -3 into the function: \[ f(-3) = (-3)^2 + 2 \times (-3) = 9 - 6 = 3 \] Therefore, \( f(-3) = 3 \).

Step 4: Evaluate f(a)

Substitute \( a \) into the function: \[ f(a) = a^2 + 2a \] Therefore, \( f(a) = a^2 + 2a \).

Step 5: Evaluate f(-x)

Substitute \( -x \) into the function: \[ f(-x) = (-x)^2 + 2(-x) = x^2 - 2x \] Therefore, \( f(-x) = x^2 - 2x \).

Step 6: Evaluate \( f\left(\frac{1}{a}\right) \)

Substitute \( \frac{1}{a} \) into the function: \[ f\left(\frac{1}{a}\right) = \left(\frac{1}{a}\right)^2 + 2\left(\frac{1}{a}\right) = \frac{1}{a^2} + \frac{2}{a} \] Therefore, \( f\left(\frac{1}{a}\right) = \frac{1}{a^2} + \frac{2}{a} \).

Key Concepts

Quadratic Functions

Quadratic functions are a type of polynomial function. They are a fundamental concept in algebra. The standard form of a quadratic function is \( f(x) = ax^2 + bx + c \). Here, \( a \), \( b \), and \( c \) are constants, and \( a \) is not equal to zero. Quadratics form a curved shape on a graph, known as a parabola, which can open either upwards or downwards. This direction depends on the sign of \( a \).

• If \( a > 0 \), the parabola opens upwards.
• If \( a < 0 \), the parabola opens downwards.

Present in a variety of situations, quadratic functions can describe trajectories, areas, and other real-world applications.
In our problem, the function \( f(x) = x^2 + 2x \) is a quadratic function because it's expressed in the form of a second-degree polynomial, where \( a = 1 \), \( b = 2 \), and \( c = 0 \). Its graph is a parabola that opens upwards because the coefficient of \( x^2 \) is positive.

Substitution

Substitution is a mathematical method used to replace variables with specific values. It’s commonly used in function evaluation to determine the output of a function for particular input values. Substitution is straightforward – you simply replace each instance of the variable in the function with the number you're evaluating.
For example, to evaluate \( f(x) = x^2 + 2x \) at \( x = 3 \), substitute \( 3 \) into the function:

• Original function: \( f(x) = x^2 + 2x \)
• Substitute \( x = 3 \): \( f(3) = 3^2 + 2 \times 3 \)
• Calculate: \( f(3) = 9 + 6 = 15 \)

This shows how the function behaves at \( x = 3 \).
Substitution is helpful for checking multiple values at once, as seen in the exercise where we evaluate \( f(x) \) at different inputs such as \( 0 \), \(-3\), and \( \frac{1}{a} \). Each substitution allows us to observe the function's behavior at various points.

Polynomial Functions

Polynomial functions consist of terms that include variables raised to whole-number powers. They can vary in complexity, from simple linear functions to complex, high-degree polynomials. In general, a polynomial function is expressed as \( f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 \), where \( n \) is the highest power of \( x \), also called the degree, and \( a_n \) are constants.
Quadratic functions, like \( f(x) = x^2 + 2x \), are a special type of polynomial function where the degree is 2. This makes them the simplest form of polynomial that isn't linear, which makes them a key focus in mathematics.

• The degree of a polynomial influences its graph's shape and the number of potential roots or solutions.
• In the context of solving and evaluating polynomials, understanding each term's role helps predict the function's overall behavior.

Tackling polynomial functions often involves techniques like factoring, expanding, and simplifying expressions. Evaluating them using specific values, as done through substitution in our exercise, aids in understanding the dynamic characteristics and solutions involved with polynomials.