Math  /  Algebra

QuestionFor a parabola in general form, f(x)=ax2+bx+cf(x)=a x^{2}+b x+c, where a0a \neq 0. If a<0a<0 then the range of f(x)f(x) is

Studdy Solution
Since the parabola opens downward (because a<0a<0), the range of f(x)f(x) is all values less than or equal to the y-coordinate of the vertex:
Range(f(x))=(,b24a+c] \text{Range}(f(x)) = \left( -\infty, \frac{-b^2}{4a} + c \right]

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