How do you write the vertex form equation of the parabola #y = x^2 + 16x + 71#?

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Answer 1 · 2018-05-10T11:51:57

Vertex form of equation is #y = ( x+8)^2 +7#

# y = x^2 +16 x +71 # or

# y = (x^2 +16 x +64) +7 # or

#y = ( x+8)^2 +7# Comparing with vertex form of

equation #f(x) = a(x-h)^2+k ; (h,k)# being vertex we find

here #h= -8 , k=7 :.# Vertex is at #(-8,7) # and vertex form of

equation is #y = ( x+8)^2 +7#

graph{x^2+16 x+71 [-40, 40, -20, 20]} [Ans]

Answer 2 · 2018-05-10T14:48:15

#y = (x+8)^2 +7#

The vertex is at #(-8,7)#

Vertex form is obtained by the process of completing the square,

#y = x^2+16x+71#

#y = x^2 +16x color(blue)(+64-64) +71" "larrcolor(blue)( +(b/2)^2 -(b/2)^2)#

#y = (x^2 +16x +64) + (-64 +71)#

#y = (x+8)^2 +7" "larr# vertex form.

The vertex is at #(-8,7)# graph{y =x^2+16x+71 [-71, 89, 7.1, 87.1]}