# x^2-2/3 x - 8/9 ; a= 1 , b= -2/3 , c= -8/9[ax^2+bx+c]#
Vertex ( x coordinate) is # v_x= (-b)/(2 a)=(2/3)/2=1/3 #
Putting #x=1/3# in the equation we get #v_y#
Vertex ( y coordinate) is # v_y= 1/9-2/9-8/9=-1#
Therefore, vertex is at #(1/3 , -1)#
Axis of symmetry is # x = 1/3 #
Since #a>0# , parabola opens upward.
x-intercepts are found by putting #y=0# in the equation
#x^2-2/3 x - 8/9=0 or 9 x^2 -6 x -8=0 # or
#(3 x+2)(3 x- 4)=0:. x = -2/3 , x = 4/3#
y-intercept is found by putting #x=0# in the equation
#y=x^2-2/3 x - 8/9 or y = -8/9#
Additional points for graph: # x=4/3 :.y=3 #
and #x = -5/3 :. y=3 and x =13/3 :.y=15 # and
#x = -11/3 , y=15# Therefore, additional points on
parabola are #(-5/3,3 and 4/3,3) # and
#(-11/3,15 and 13/3,15) #
graph{x^2-2/3 x-8/9 [-20, 20, -10, 10]} [Ans]
