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CLEP College Algebra outcomes include analyzing quadratic functions using vertex form and completing the square. For f(x)=x2−6x+8f(x)=x^2-6x+8, complete the square: x2−6x=(x−3)2−9x^2-6x=(x-3)^2-9. Thus f(x)=(x−3)2−9+8=(x−3)2−1f(x)=(x-3)^2-9+8=(x-3)^2-1. A perfect-square term (x−3)2(x-3)^2 is always ≥0\ge 0, so the smallest value occurs when it equals 0, at x=3x=3. Substituting x=3x=3 gives f(3)=−1f(3)=-1. Graphically, the parabola opens upward (leading coefficient >0>0) and its vertex at (3,−1)(3,-1) is the minimum point. Alternative methods—using −b/(2a)-b/(2a) to find the vertex’s x-coordinate—yield the same result: with a=1,b=−6a=1, b=-6, x=−(−6)/(2⋅1)=3x=-(-6)/(2\cdot1)=3, then f(3)=−1f(3)=-1. On CLEP algebra items, recognizing vertex form streamlines questions about extrema, axis of symmetry, and transformations. Also note that choosing options like 11 or −5-5 often reflects arithmetic slips (e.g., miscomputing constant terms after completing the square). Mastery of these standard techniques helps with function analysis, optimization, and interpreting quadratic models across contexts.