What is the vertex form of #y= 8x^2+17x+1 #?
1 Answer
Jan 26, 2016
# y =8 (x + 17/16 )^2 - 257/32 #
Explanation:
The vertex form of the trinomial is;
#y = a(x - h )^2 + k# where (h , k ) are the coordinates of the vertex.
the x-coordinate of the vertex is x
# = -b/(2a)# [from
# 8x^2 + 17x + 1 # a = 8 , b = 17 and c = 1 ]
so x-coord
# = -17/16 # and y-coord
# = 8 xx (-17/16)^2 + 17 xx (-17/16) + 1 #
# = cancel(8) xx 289/cancel(256) - 289/16 + 1#
# = 289/32 - 578/32 + 32/32 = -257/32# Require a point to find a: if x = 0 then y=1 ie (0,1)
and so : 1= a
#(17/16)^2 -257/32 = (289a)/256 -257/32# hence
# a = (256 + 2056)/289 = 8# equation is :
# y = 8( x + 17/16)^2 - 257/32#
