The graph of y+x^2 =0 lies in which quadrants?

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Answer 1 · 2018-04-22T09:09:31

The graph of #y+x^2=0# lies in #Q3# and #Q4#.

#y+x^2=0# means that #y=-x^2# and as whether #x# is positive or negative, #x^2# is always positive and hence #y# is negative.

Hence the graph of #y+x^2=0# lies in #Q3# and #Q4#.

graph{y+x^2=0 [-9.71, 10.29, -6.76, 3.24]}

Answer 2 · 2018-04-22T09:36:22

Quadrants 3 and 4.

To solve this equation, the first step would be to simplify the equation #y+x^2=0# by isolating #y# as follows:

#y+x^2 = 0#

#y+x^2-x^2 = 0-x^2#

To isolate #y#, we subtracted #x^2# from both sides of the equation.

This means that #y# can never be a positive number, only #0# or a negative number, since we stated that #y# equals a negative value; #-x^2#.

Now to graph it out:
graph{y=-x^2 [-19.92, 20.08, -16.8, 3.2]}
We can test that the graph is correct simply by using a value for #x#:

#x=2#

#y=-(2^2)#

#y=-4#

If you zoom in on the graph, you can see that when #x=2#, #y=-4#.

Because the graph is symmetrical, when #y=-4#, #x=2 or x=-2#.

And to answer your question, we can see that when we plot the equation out on the graph, the line falls in quadrants 3 and 4.