Given: #f(x) = x^2 + 6x + 3#
When the equation is in the form: #Ax^2 + Bx + C #,
the vertex is #(-B/(2A), f(-B/(2A)))#
Find the vertex:
#x = -B/(2A) = -6/2 = -3#
#f(-3) = (-3)^2 + 6(-3) + 3 = 9 -18 + 3 = -6#
vertex #(-3, -6)#
Find y-intercept, by setting #x# = 0#
#y = f(0) = 0^2 + 6*0 + 3 = 3 => (0, 3)#
Find the x-intercepts , by setting #y = 0#:
#0 = x^2 + 6x + 3#
Use the quadratic formula: #x = (-B+- sqrt(B^2 - 4 A C))/(2A)#
#x = -6+- sqrt(36 - 4(1)(3))/2 = -3 +- sqrt(24)/2#
#x = -3 +- (sqrt(4)sqrt(6))/2 = -3 +- (2sqrt(6))/2#
#x = -3 +- sqrt(6)#
#x#-intercepts #(-3 - sqrt(6), 0), (-3 + sqrt(6), 0)#
Find additional points:
Since #x# is the independent variable, we cal select any #x# and solve for the corresponding #y# to find more points:
#ul(" "x " "|" "y" ")#
#-1" "| -2"#
#-2" "| -5"#
#-4" "| -5"#
#-6" "| -3"#
