What is the vertex of #y=x^2+6x+5#?

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Answer 1 · 2016-01-05T14:57:53

( -3 , -4)

making use of the standard form of a trinomial ie. # ax^2 + bx + c #

for y = # x^2 + 6x + 5 #

a = 1 , b = 6 and c = 5

the x-coordinate of the vertex = - (b/2a)

# rArr x = - 6/2 # = - 3

now substitute this value of x into the equation to obtain corresponding value of y.

# rArr y = ( -3 )^2 + 6( -3 ) + 5 # = 9 - 18 + 5 = - 4
graph{x^2+6x+5 [-10, 10, -5, 5]}