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In mathematics, substitution helps us evaluate functions by replacing variables with specific values. This process allows us to determine the output of a function when given a particular input. For example, let's look at the function \(f(x)=x^{2}+6x+9\). This function has an input variable, \(x\), which we can replace to find specific results.

To solve the exercise, we are asked to find \(f(-3)\). This involves substituting \(-3\) for every \(x\) in the function. This step-by-step replacement looks like this: replace \(x\) in \(x^{2}\) with \(-3\), resulting in \((-3)^{2}\), and do the same for the term \(6x\), making it \(6(-3)\). Finally, include \(9\) as a constant in the expression.

• First term: \((-3)^{2} = 9\)
• Second term: \(6(-3) = -18\)
• Constant term: \(9\)

Once the substitution is complete, we can then evaluate the entire function to find the result.

Evaluating functions means finding the output value for a given input by substituting the input into the given function. After substitution, the next step is to perform the necessary arithmetic operations to simplify the expression. Let's continue from where we substituted \(-3\) into the function \(f(x) = x^{2} + 6x + 9\).

After substitution, the function becomes \(f(-3) = 9 - 18 + 9\). Now we start calculating:

• Calculate \(9 - 18 = -9\)
• Add \(9\) to \(-9\) to simplify the expression to \(-9 + 9 = 0\)

Thus, the function evaluates to \(f(-3) = 0\). Evaluating functions consistently involves careful substitution, operation execution, and simplification.

A polynomial function is a mathematical expression consisting of variables, coefficients, and exponents arranged in a sum of terms. The generic form of a polynomial function can be \(a_nx^n + a_{n-1}x^{n-1} + \, ... + \, a_1x + a_0\), where \(a_n\) to \(a_0\) are constants and \(n\) is a non-negative integer.

In our exercise, the function \(f(x) = x^2 + 6x + 9\) is a polynomial. Specifically, it is a quadratic polynomial because the highest degree exponent on \(x\) is \(2\). Understanding the structure of polynomial functions is crucial because:

• They form the basis for much of algebra.
• They are easily solvable using known methods such as factoring.
• They represent a wide class of problems in mathematics.

Recognizing that our function simplifies to a linear expression \(9 - 18 + 9 = 0\) after evaluation shows the simplicity hidden in polynomial structures when evaluated correctly.