What is the vertex form of #y= x^2 + 5x + 6 #?

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Answer 1 · 2017-07-24T03:00:56

Vertex form is #(x+5/2)^2-1/4#.

Vertex from Standard Form

#y=x^2+5x+6# is the standard form for a quadratic equation, #ax^2+bx+6#, where #a=1#, #b=5#, and #c=6#.

The vertex form is #a(x-h)^2+k#, and the vertex is #(h,k)#.

In the standard form, #h=(-b)/(2a)#, and #k=f(h)#.

Solve for #h# and #k#.

#h=(-5)/(2*1)#

#h=-5/2#

Now plug in #-5/2# for #x# in the standard form to find #k#.

#f(h)=k=(-5/2)^2+(5xx-5/2)+6#

Solve.

#f(h)=k=25/4-25/2+6#

The LCD is 4.

Multiply each fraction by an equivalent fraction to make all of the denominators #4#. Reminder: #6=6/1#

#f(h)=k=25/4-(25/2xx2/2)+(6/1xx4/4)#

Simplify.

#f(h)=k=25/4-50/4+24/4#

Simplify.

#f(h)=k=-1/4#

Vertex #(-5/2,-1/2)#

Vertex form: #a(x-h)^2+k#

#1(x+5/2)^2-1/4#

#(x+5/2)^2-1/4#