How do you find the intercepts, vertex and graph #f(x)=x^2-2/3x-8/9#?

1 Answer
Aug 11, 2018

Vertex: #(1/3,-1)#, symmetry: #x=1/3# , y intercept: #y=-8/9#, x intercepts : #x=-2/3 , x=4/3#,additional points: #(-5/3,3 and 4/3,3)# and #(-11/3,15 and 13/3,15) #

Explanation:

# 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]