What is the vertex form of #y=8x^2 + 19x + 12 #?

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Answer 1 · 2016-10-16T21:57:11

#y = 8(x - -19/16)^2 + 23/32#

The equation is in the standard form, #y = ax^2 + bx + c# where #a = 8, b = 19, and c = 12#

The x coordinate, h, of the vertex is:

#h = -b/(2a)#

#h = -19/(2(8)) = -19/16#

To find the y coordinate, k, of the vertex, evaluate the function at the value of h:

#k = 8(-19/16)(-19/16) + 19(-19/16) + 12#

#k = (1/2)(-19)(-19/16) + 19(-19/16) + 12#

#k = - 19^2/32 + 12#

#k = - 361/32 + 12#

#k = - 361/32 + 384/32#

#k = 23/32#

The vertex form of the equation of a parabola is:

#y = a(x - h)^2 + k#

Substitute our values into that form:

#y = 8(x - -19/16)^2 + 23/32#