Given: #y = x^2 + 2x -3#

**Find the vertex** : When #y= Ax^2 + Bx + C = 0# the vertex is #(h, k)#, where #h = (-B)/(2A)#.

#h = (-B)/(2A) = -2/(2*1) = -1#

#k = f(h) = (-1)^2 + 2(-1) - 3 = -4#

#"vertex":(-1, -4)#

**Find #x#-intercepts** by setting #y = 0# and by factoring: #" "y = (x - 1)( x + 3) = 0#

#x"-intercepts": (1, 0), (-3, 0)#

**Find the #y#-intercept** by setting #x = 0#:

#y = --3; " "y"-intercept":(0, -3)#

**Find Domain and range** :

Domain is all the valid input #(x)#. Since there is no radical such as a square root, or no denominator that could become undefined, the domain is all valid Reals.

Domain: all Reals, #" "-oo < x , oo " or " (-oo, oo)#

Range is limited by the input. The lowest #y#-value is the vertex.

Range: #" "-4 <= y < oo " or " #[-4, oo)#