What is the vertex, axis of symmetry, the maximum or minimum value, and the range of parabola #y = –3(x + 8)^2 + 5#?

Question

Savannah Anderson · Accepted Answer

1) #(-8,5)#
2) #x=-8#
3) max = #5#, min = #-infty#
4) R = #(-infty,5]# 1) let's traslate: #y'=y# #x'=x-8# so the new parabola is #y'=-3x'^2+5# the vertex of this parabola is in #(0,5) =># the vertex of the old parabola is in #(-8,5)# NB: you could have solve this even without the translation, but it would have been just a waste of time and energy :) 2) The axis of symmetry is the vertical lie passing through the vertex, so #x=-8# 3) It is a downward-facing parabola because the directive coefficient of the quadratic polynomial is negative, so the max is in the vertex, i.e. max=5, and the minimum is #-infty# 4) Because it is a continuous function, it satisfies Darboux property so the range is #(-infty,5]# NB: If you don't know Darboux property, it is trivial to prove that if #exists y_0 < y_1 : exists x_0 and x_1 : y_0=-3(x_0+8)^2+5# and #y_1=-3(x_0+8)^2+5#, so #forall y in (y_0,y_1) exists x : y=-3(x+8)^2+5#, you just have to solve the equation and use the relations to prove that #Delta>=0#

Eva Adamson · Answer

undefined Vertex: (-8, 5)
Axis of symmetry: x = -8
Maximum value: 5
Range: y ≤ 5