What are the important points needed to graph # y = 3( x + 1 )^2 -4#?

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Answer 1 · 2018-06-22T17:51:04

see graph.

this is in vertex form:

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

the vertex is #(-h,k)#

Axis of symmetry #aos=-h#

#a>0# open up, has a minimum

#a<0# opens down has a maximum

you have:

vertex #(-1,-4)

#aos =-1#

set #x=0# to solve y-intercept:

#y = 3( x + 1 )^2 -4#

#y = 3( 0 + 1 )^2 -4 = -1#

#y=-1#

set #y=0# to solve x-intercept(s) if they exist:

#y = 3( x + 1 )^2 -4#

#0= 3( x + 1 )^2 -4#

#4/3 = ( x + 1 )^2#

#+-sqrt(4/3) = x + 1#

#x=-1+-sqrt(4/3)#

#a=5# so #a>0# parabola opens up and has a minimum at vertex.

graph{3( x + 1 )^2 -4 [-10, 10, -5, 5]}