1 Introduction to Evaluate the function \( f(x)=x^{2}-2 x+1 \) at the given values of the independent variable and simplify. \( \begin{array}{lll}\text { a. } f(-2) & \text { b. } f(x+6) & \text { c. } f(-x) \\ \text { Part } 1 \text { of } 3\end{array} \) \( \begin{array}{l}\text { a. } f(-2)=\square \text { (Simplify your answer.) }\end{array} \)

To evaluate the function \( f(x) = x^2 - 2x + 1 \) at \( x = -2 \), simply substitute -2 into the function: \[ f(-2) = (-2)^2 - 2(-2) + 1 = 4 + 4 + 1 = 9. \] So, \( f(-2) = 9 \). Now onto the fun part! Did you know this quadratic function can be visualized as a parabola opening upwards? It has its vertex at the point (1, 0), which is the minimum value of the function. For part b, we substitute \( x + 6 \) into the function: \[ f(x + 6) = (x + 6)^2 - 2(x + 6) + 1. \] Expanding it out gives: \[ (x^2 + 12x + 36 - 2x - 12 + 1) = x^2 + 10x + 25. \] So, \( f(x + 6) = x^2 + 10x + 25 \). If you recognize this, it’s actually a square trinomial: \( (x + 5)^2 \)! Check this out: the function’s evaluations have a unique pattern, revealing the elegance of algebra!