#f(x) =1/4(x+2)^2-9#
#= 1/4(x^2+4x+4)-9#
#= x^2/4 +x -8#
First let's find the zeros of #f(x)#
#x^2/4 +x -8 =0#
#x^2 +4x -32 =0#
#(x+8)(x-4)=0#
#:. x= -8 or +4# are the zeros of #f(x)#
#:.# the #x-# intercepts of #f(x)# are the points #(-8,0)# and #(4,0)#
Next, let's find the extrema of #f(x)#
#f(x)# is a quadratic of the form #ax^2+bx+c#.
The graph of #f(x)# is a parabola with vertex at #x=(-b)/(2a)#
#x_"vertex" = (-1)/(2*(1/4))=-2#
Since #a=1/4>0# the #f(x_"vertex" )# is the absolute minimum of #f(x)#
#:. f_min = f(-2) = 4/4-2-8= -9#
Hence, the absolute miminum of #f(x)# is the point #(-2,-9)#
Finally, let's find the #y-#intercept of #f(x)#
#f(0) = -8#
Hence, the #y-#intercept of #f(x)# is the point #(0,-8)#
We now have the critical points to graph #f(x)# below.
graph{1/4(x+2)^2-9 [-28.87, 28.86, -14.43, 14.44]}
