Answers needed.

Question

Consider the function f(x) = -x^2-12x - 38.

1. What is the vertex of the graph of this function? _______  

1. The maximum/minimum value is  _______.  

The axis of symmetry is  _______.

Answer 1

y=-x^2 -12x -38

y' =--2x-12=0

-2x = 12

x = 12/-2 =-6

axis of symmetry is x=-6

vertex or maximum point has x coordinate of -6, plug that into the y equation to get the y coordinate of the vertex. y=-(-6)^2 -12(-6) -38 = -36+72-38 = -2. vertex is (-6-2)= maximum point. It's a downward opening parabola.

-x^2-12x -38 =0

x^2 +12x +38 =0

x^2 +12x + (12/2)^2 = -38 +(12/2)^2

(x+6)^2 = -38 +36 =-2

x-6 =sqr(-2)

x = 6 + or - isqr2 (there are no x intercepts. the parabola is below the x axis and never intersects the x axis