How do you find the range of #f(x)= -6(x+4)^2-12#?

Question

Eva Adamson · Accepted Answer

#(-oo,-12]# #"we require to find the vertex and if max/min"# #"the equation of a parabola in "color(blue)"vertex form"# is. #color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+b)color(white)(2/2)|)))# #"where "(h,k)" are the coordinates of the vertex and a"# #"is a multiplier"# #y=-6(x+4)^2-12" is in vertex form"# #"with vertex "=(-4,-12)# #"to determine if vertex is max/min"# #• " if "a>0" then minimum "uuu# #• " if "a<0" then maximum "nnn# #"here "a=-6" hence maximum at "(-4,-12)# #"range is "(-oo,-12]# graph{-6(x+4)^2-12 [-40, 40, -20, 20]}

Eli Assouline · Answer

undefined To find the range of $ f(x) = -6(x+4)^2 - 12 $, we can analyze the behavior of the function. Since the function is a quadratic in the form $ ax^2 + bx + c $, where $ a $ is negative, the graph of the function opens downwards. Therefore, the maximum value of the function occurs at the vertex.

The vertex of the parabola represented by $ f(x) = -6(x+4)^2 - 12 $ is at the point $ (-4, -12) $.

Since the parabola opens downwards, the range of the function is all real numbers less than or equal to the y-coordinate of the vertex.

So, the range of $ f(x) $ is $ \{ y \in \mathbb{R} \, | \, y \leq -12 \} $.