What is the vertex form of #y=3x^2+2x-8 #?

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Answer 1 · 2017-10-07T06:10:14

#y=3(x+0.bar(3))^2-8.bar(3)#

Vertex form is written:

#y=a(x-h)^2+k#

Where #(h,k)# is the vertex.

Currently the equation is in standard form, or:

#y=ax^2+bx+c#

Where #(-b/(2a),f(-b/(2a)))# is the vertex.

Let’s find the vertex of your equation:

#a=3 and b=2#

So,

#-b/(2a)=-2/(2*3)=-2/6=-1/3#

Thus #h=-1/3=-0.bar(3)#

#f(-1/3)=3(-1/3)^2+2(-1/3)-8#
#f(-1/3)=3(1/9)-2/3-8#
#f(-1/3)=1/3-2/3-8=-8.bar(3)#

Thus #k=-8.bar(3)#

We already know that #a=3#, so our equation in vertex form is:

#y=3(x-(-0.bar(3)))^2+(-8.bar(3))#

#y=3(x+0.bar(3))^2-8.bar(3)#