How do you find the y intercept, axis of symmetry and the vertex to graph the function #f(x)=x^2-2/3x-8/9#?

1 Answer
Jan 2, 2017

#y# intercept: at #y=-8/9#
Axis of symmetry: #x=1/3#
Vertex: #(1/3,-1)#

Explanation:

y-intercept
The y-intercept is the value of #y# when #x=0#.
Substituting #0# for #x# in #y=x^2-2/3x-8/9#
gives #y=-8/9#

Axis of symmetry
For a parabola in the form #y=x^2+px+q#
the axis of symmetry is #x=p/2#
For the given example #p=-2/3#
so #p/2 =-1/3#
(see bottom for alternate derivation)

Vertex
Given an axis of symmetry #x=t#
The vertex is at #(a,f(t))#
In this case #f(-1/3) = (-1/3)^2-(-1/3) * 2/3 -8/9 = 1/9-2/9-8/9=-1#

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Alternate Deviation of Axis of Symmetry and Vertex
A parabola in the form #y=(x-a)^2+b#
has a vertex at #(a,b)#
and an axis of symmetry #x=a#

#y=2^2-2/3x-8/9# can be converted into this form by "completing the square".

#y=x^2-2/3xcolor(red)(+(-1/3)^2)-8/9color(red)(-(1/3)^2)#

#color(white)("X")=(x-1/3)^2-1#
giving:
#color(white)("XXX")#vertex at #(1/3,-1)# and
#color(white)("XXX")#axis of symmetry #x=1/3#

graph{x^2-2/3x-8/9 [-2.947, 3.215, -2.12, 0.958]}