How do you find the vertex of #y = 4x^2 + 8x + 7#?

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Answer 1 · 2016-03-28T07:52:14

#(h, k)=(-1, 3)#

From the given equation

#y=4x^2+8x+7#

There is a formula for finding the vertex #(h, k)# for equation of the parabola of the form #y=ax^2+bx+c#

#h=-b/(2a)# and #k=c-b^2/(4a)#

Solve for #h# with #a=4# and #b=8# and #c=7#

#h=-b/(2a)=(-8)/(2*4)=-1#

Solve for #k#

#k=c-b^2/(4a)=7-8^2/(4*4)=7-64/16=7-4=3#

Our vertex is at #(-1, 3)#

See the graph of #y=4x^2+8x+7# with vertex at #(-1, 3)#
graph{y=4x^2+8x+7[-20,20,-10,10]}

God bless....I hope the explanation is useful.