What is the axis of symmetry and vertex for the graph # y= -x^2-4x+5#?

There are different ways of going about this:

#1.# Solve y=0 to find the #x#-intercepts
#y=-x^2-4x+5#
#0=-x^2-4x+5#
#0=(x+5)(-x+1)#

Therefore #x = -5 and x = 1# are the #x#-intercepts.

The axis of symmetry has to be between these two points:

-5, -4, -3, -2, -1, 0, 1 #" "# Or #x = (-5+1)/2 = -4/2 = -2#

Hence, our axis of symmetry is #x=-2#.

Because we know the graph will be at its lowest or highest (i.e. its vertex), on the line #x=-2#, we can plug #-2# into the original equation and find out the vertex point:

#y =-(-2)^2 -4(-2)+5#

#y=9#

Therefore the vertex is at #(-2,9)#

#2#. Complete the square
This is an alternative method, which is very commonly used in middle and high school. I will not get into how to complete the square, and just show you what I did:

#y=-x^2-4x+5#
#y=-(x^2+4x-5)#
#y =-(x^2 +4x +4-4-5)#
#y =-[(x+2)^2-9)]#

#y=-(x+ **2** )^2 **+9** #

Therefore the vertex is at (#-2,9)#

The problem requires the given quadratic function, to be put in vertex form as demonstrated below. The quadratic function represents a vertical parabola.

#y= -(x^2 +4x)+5#

=#-(x^2 +4x +4 -4)+5#

=#-(x^2 +4x+4) +4+5#

=#-(x+2)^2 +9#

The given quadratic function is now in vertex form. This represents a vertical parabola, opening down wards, with its vertex at (-2,9) and the axis of symmetry being x=-2