Question

Which quadratic function has a range that can be described as y≤10.5
?(1 point)

f(x)=−6(x+4)2+10.5


f(x)=10.5(x−4)2−6


f(x)=−6(x−10.5)2+4


f(x)=6(x+4)2+10.5

Answer 1

f(x) = −6(x+4)^2 + 10.5

Reason: This is in vertex form with vertex y = 10.5 and a = −6 < 0, so the parabola opens downward and its range is y ≤ 10.5.