What is the vertex, axis of symmetry, the maximum or minimum value, and the range of parabola #f(x) = −4(x − 8)^2 + 3#?

Question

Abigail Allen · Accepted Answer

#f(x)=-4(x-8)^2+3#
is a standard quadratic in vertex form:
#f(x) = m(x-a)^2+b#
where #(a,b)# is the vertex. The fact that #m=-4<0# indicates that the parabola opens downward (the vertex is a maximum value) The vertex is at #(8,3)# Since it is a standard position parabola, the axis of symmetry is
#x=8# The maximum value is #3# The Range of #f(x)# is #(-oo,+3]#

Abigail Allen · Answer

undefined Vertex: The vertex of the parabola $ f(x) = -4(x - 8)^2 + 3 $ is at the point (8, 3).
Axis of Symmetry: The axis of symmetry of the parabola is the vertical line passing through the vertex. So, the axis of symmetry is $ x = 8 $.
Maximum or Minimum Value: Since the coefficient of the $ x^2 $ term is negative, the parabola opens downward. Therefore, the maximum value occurs at the vertex. In this case, the maximum value is 3.
Range: Since the parabola opens downward and the maximum value is 3, the range of the function is $ (-\infty, 3] $.