Since the #x^2# term of the quadratic has a positive coefficient we know the parabola is of the #uuu# form.
From the discriminant we know that if:
#sqrt(b^2-4ac)=0# the parabola turns at the x axis, resulting in there being one value for which y = 0.
If:
#sqrt(b^2-4ac)>0# the parabola will cross the x axis resulting in there being values of x such that y < 0.
But if:
#sqrt(b^2-4ac)<0# then the parabola will be above the x axis and so the function will be greater than zero for all values of x. This is therefore what we need to solve to find the required values of h.
#b^2 = (h-3)^2#
#:.#
#(h-3)^2-4(1)(4)<0#
#h^2-6h-7=0#
Factoring:
#(h-7)(h+1)=0=> h=7 , h= -1#
These are the boundary points. We are looking for values less than zero, so these will be when:
#h<7# and #h> -1#
Hence:
#-1 < h < 7#
