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

Answer 1

#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 approach:

#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#
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Answer 2

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)

When we compare the two equations, the vertex is (-4,4).

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

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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