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]}

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