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Polynomial functions are a fundamental building block in mathematics. They consist of terms that are non-negative integer powers of the variable, typically written as coefficients multiplied by powers of the variable. The general form looks like this:\[f(x) = a_n x^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\]Where:

• \(a_n, a_{n-1}, \ldots, a_0\) are coefficients.
• \(x\) is the variable.
• \(n\) is a non-negative integer that dictates the degree of the polynomial.

In our exercise, the polynomial function is \(f(x) = 5x^2 - 6x + 1\). It's a quadratic polynomial because it has a degree of 2, as indicated by the highest power of \(x\), which is 2. Understanding the structure of polynomial functions is crucial as it helps in performing operations like addition, subtraction, and differentiation, which are frequently required in calculus and real-world applications. This structured form makes it easier to manipulate and calculate within mathematics.

The substitution method is an essential technique used to find the value of a function for a given variable. It involves replacing the variable in the function with another expression or specific value. In our exercise, we are asked to substitute \(x\) and \(x+h\) into the function.Here's how it works:

• Substitute \(x+h\) into the polynomial function to find \(f(x+h)\).
• Retain the existing term of \(f(x)\) for comparison.

So, substituting \(x+h\) into the polynomial \(f(x)\), we get:\[f(x+h) = 5(x+h)^2 - 6(x+h) + 1\]Expanding this:

• \(5(x+h)^2 = 5(x^2 + 2hx + h^2) = 5x^2 + 10hx + 5h^2\)
• \(-6(x+h) = -6x - 6h\)
• The constant term remains \(+1\).

This yields \[f(x+h) = 5x^2 + 10hx + 5h^2 - 6x - 6h + 1\]Differentiating and working with these substitutions ensures you're assessing how small changes in \(x\) (namely, \(h\)) affect the function's value. It's a foundational skill for calculating derivatives, an integral part of calculus.

Simplifying expressions is a critical skill in algebra and calculus. It's about reducing a mathematical expression to its simplest form, making it easier to understand and work with.For our specific purpose, the task was to simplify the expanded polynomial expressions derived from substitution:

• Start by expanding \(f(x+h) = 5x^2 + 10hx + 5h^2 - 6x - 6h + 1\).
• Then, subtract \(f(x) = 5x^2 - 6x + 1\) from \(f(x+h)\).

After the subtraction:\[f(x+h) - f(x) = (5x^2 + 10hx + 5h^2 - 6x - 6h + 1) - (5x^2 - 6x + 1)\]The expression simplifies to:

• The \(5x^2\) and \(-6x\) terms cancel out since they are in both expressions.
• The constant term \(+1\) likewise cancels out.
• You are left with \(10hx + 5h^2 - 6h\).

From here, finding the difference quotient as:\[\frac{f(x+h) - f(x)}{h} = 10x + 5h - 6\]This simplification makes it easy to interpret how the function changes in response to small changes in \(x\). Mastering this skill paves the way for more complex calculus concepts like limits and derivatives.