What is the vertex of # y=4/3x^2 - 2x - 3 #?

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Answer 1 · 2017-11-16T02:05:30

#Vertex(3/4,-15/4)#

In this form of the Parabola equation, i.e.:

#ax^2+bx+c#

the vertex has coordinates of:

#x=-b/(2a)# and #y=f(-b/(2a))#

In this problem:

#a=4/3# and #b=-2# and #c=-3#

#x#-coordinate of the vertex =#(-(-2))/(2(4/3))=2/(8/3)=2*(3/8)=3/4#

#y#-coordinate of the vertex can be found by plugging in the value of the #x#-coordinate into the equation of the Parabola.

#y=(4/3)(3/4)^2-2(3/4)-3#

#y=(4/3)(9/16)-(3/2)-3#

#y=3/4-3/2-3#

#y=(3-6-12)/4=-15/4#