When graphing y = x2 – 14x + 24 and y = 20 – |x + 3|, which value would be the best minimum x- and minimum y-values for the viewing window so that all intercepts, minimums, maximums, and points of intersection are able to be seen as closely as possible?

The parabola has its minimum at (7, −25), since its vertex is found by completing the square: y = (x − 7)² − 25. The absolute-value graph has x-intercepts at −23 and 17, so the window’s minimum x-value must be no greater than −23. A minimum of −25 is the closest listed value that includes both the leftmost intercept and the parabola’s minimum, so choice C, −25, is best.

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