help

czagara

Junior Member
Joined
Sep 24, 2009
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67
Sketch the graph of the equation. Identify any intercepts and test for symmetry. (Plot the intercepts and the graph.)

y = -x^2-2x

I came up with no symmetry and the intercepts (-2,0) and (0,0) am i right?
 
Well, after plotting the graph -- yes you are right, \(\displaystyle (-2,0)\) and \(\displaystyle (0,0)\) are the points where this graph intercepts the x - axis.

However, about the symmetry, look a bit closer. Doesn't this function seem to be reflected across the line \(\displaystyle x = -1\)?
 
i am having trouble trying to figure out what my graph should look like. I am trying to sketch it on an online class and its not taking this as the answer but it did take no symmetry
 
JuicyBurger is correct. There is symmetry about the vertical line x = -1.

With respect to being an even function (which it is not), there is no symmetry about the y-axis. Is that what your machine is talking about?
 
no i entered no symmetry in the box and submitted it and it was correct. It did give me the option for the x-axis and y-axis symmetry but like i said i submitted no symmetry and the answer was correct.
 
If the machine "thinks" that no symmetry means only no axial symmetry, then its response is correct. Otherwise, it's wrong.
 
ok but still i need help with the graph, its not accepting those points. I think its looking for a parabola of some sort. Any suggestions?
 
The on-line class provides you with the format it expects to receive, yes? What specifically is it asking for?
 
Sketch the graph of the equation. Identify any intercepts and test for symmetry. (Plot the intercepts and the graph.)
y = ?x2 ? 2x

That is the exact question verbatim and then there is a graph underneathe it and it wants me to do exactly what the question states. I have already done the symmetry part and now I need the graph part.
 
i have (-2,0) on my graph and (0,0) on my graph and its not accepting.
 
Two points do not define a parabola.

How about you determine the coordinates of the vertex, and plot that, too.

The vertex lies on the parabola's axis of symmetry, which we've already seen to be x = -1.

What is y when x is -1 ?

Plot that point.
 
Yes,

\(\displaystyle -(-1)^2-2(-1) = 1\), therefore \(\displaystyle y(-1) =1\) and your point will be \(\displaystyle (-1,1)\) for the vertex of the parabola
 
Unless you tell me about the instructions that the on-line provider gave you for how to enter graphs, I don't think that I have any more suggestions.
 
I am with mmm4444bot on this one, I'm not quite sure what your online submission is asking for.

My best suggestion would be to find a few more points by plugging in larger values of x and then plotting those as well as the 3 we have already determined.
 
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