What is the vertex form of #y=x^2+4x+16#?

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Answer 1 · 2016-01-09T15:58:56

#y = (x + 2 )^2 + 12#

The standard form of a quadratic equation is:

#y = ax^2 + bx + c#

The vertex form is : # y = (x - h )^2 + k# where (h , k ) are the coordinates of the vertex.

For the given function #a = 1# , #b = 4#, and #c = 16#.

The x-coordinate of the vertex (h) # = -b/(2a) = - 4/2 = - 2 #

and the corresponding y-coordinate is found by substituting x =- 2 into the equation :

#rArr y = (- 2 )^2+4(- 2 ) + 16 = 4 - 8 + 16 = 12 #

the coordinates of the vertex are (- 2 , 12 ) = (h , k )

the vertex form of # y = x^2 + 4x + 16# is then :

# y = (x + 2 )^2 + 12 #

check:

#(x + 2 )^2 + 12 = x^2 + 4x +16#