What is the vertex form of #y= x^2+8x+20 #?

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Answer 1 · 2018-03-11T01:20:51

Vertex is (-4,4 )

#y=x^2+8x+20 #

this can also be written as ,
y = #x^2 + 8x + 4^2 - 4^2 + 20#

which can be further simplified into,
y = #(x+4)^2 + 4# ........ (1)

We know that,
#y = (x-h)^2 + k # where vertex is (h,k)

comparing both the equations we get vertex as (-4,4)

graph{x^2 + 8x +20 [-13.04, 6.96, -1.36, 8.64]}

Answer 2 · 2018-03-11T01:31:54

#y=(x+4)^2 +4#

The vertex form is: #y=a(x-h)^2+k#

when #(h, k)# is dhe vertex of the parabola #ax^2+bx+c#

#h=-b/(2a)#, #k=-Delta/(4a)=-(b^2-4ac)/(4a)#.

Now: #y=x^2+8x+20rArrh=-8/2 =-4# and #k=-(64-4*1*20)/(4*1)=4#

then the vertex form is: #y=(x+4)^2 +4#

Second method:

#y=x^2+8x+20rArr y-20=x^2+8xrArr#

#y-20+16=x^2+8x +16rArr y-4=(x+4)^2rArr#

#y=(x+4)^2 +4#