How do you find the vertex and the intercepts for #y=x^2+6x-5#?

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Answer 1 · 2016-03-13T01:45:39

vertex #(-3, -14)#
Y- Intercept #(0, -5)#

Given -

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

Find the vertex

#x=(-b)/(2a)=(-6)/(2 xx1)=-6/2=-3#

Find #y# at #x=-3#

#y=(-3)^2+6(-3)-5#
#y=9-18-5#
#y=9-23#
#y=-14#

vertex #(-3, -14)#

Put #x=0 #to find y-intercept

#y=0^2+6(0)-5#
#y=-5#
Y- Intercept >#(0, -5)#