Question 82584 • AZ-Common Core Algebra II A
A quadratic has a maximum when it opens downward, so its leading coefficient must be negative. For q(x) = –5(x + 10)² – 1, the vertex is (–10, –1), which is left and down from (0, 0). Rewriting t(x) gives –2(x + 1)² – 1, with vertex (–1, –1), also left and down. Thus the correct choices are q(x) and t(x).
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