SOLUTION: Graphing Nonlinear Functions:
f(x)=2(x+1)^2
The only mark on the graph I could accurately come up with was 2(0+1)^2=2 for the middle point of the Parabola. When I try to cre
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-> SOLUTION: Graphing Nonlinear Functions:
f(x)=2(x+1)^2
The only mark on the graph I could accurately come up with was 2(0+1)^2=2 for the middle point of the Parabola. When I try to cre
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Question 803219: Graphing Nonlinear Functions:
f(x)=2(x+1)^2
The only mark on the graph I could accurately come up with was 2(0+1)^2=2 for the middle point of the Parabola. When I try to create symetry but putting in 1,2,-1,-2 I end up all over the place. What am I doing wrong? Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website! It looks like would be at the
vertex of the parabola because if I go equally in
the (+) and (-) directions from ,
I get the same
and
------------
and
------------
etc.
So, the vertex is at ,
Here's the plot:
