Given: #y = (x + 3)^2 + 5#
The vertex can be found using the vertex form: #y = (x - h)^2 + k#,
where the vertex is #(h, k)#
In the given equation the #color(blue)("vertex": (-3, 5))#
The #y#-intercept is found setting #x = 0#:
#y = (0 + 3)^2 + 5 = 14#
#color(blue)(y"-intercept "(0, 14))#
Find #x#-intercepts by setting #y = 0# and factor or use the quadratic formula.
Put the equation in #Ax^2 + Bx + C = 0# form. Use the squaring formula #(a + b)^2 = a^2 + 2ab + b^2#:
#y = #x^2 + 6x + 9 + 5#
#y = x^2 + 6x + 14#
Using the quadratic formula #x = (-B +- sqrt(B^2 -4AC))/(2A)#:
#x = (-6 +- sqrt(36 -4(1)(14)))/(2) = -3 +- (sqrt(-24))/2#
Since there is a negative under the radical, the #x# values are imaginary. This means there are #color(blue)("no "x"-intercepts")#.
To plot more points, you would need to do point plotting. Since #x# is the independent variable, you would select different #x# values and calculate the #y# values.
#ul(" "x " "| " "y" ")#
# " "-6" "|" "14#
#" "-4" "|" "6#
#" "-2" "|" "6#
#" "-1" "|" "9#
graph{(x + 3)^2 + 5 [-20, 10, -4, 20]}
