SOLUTION: i need help sketching a graph that has two real zeros at -2 and 2 then another that has two imaginary zeros at -2i and 2i.i have no idea how to start it

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Question 1018018: i need help sketching a graph that has two real zeros at -2 and 2 then another that has two imaginary zeros at -2i and 2i.i have no idea how to start it
Answer by josgarithmetic(39632) About Me  (Show Source):
You can put this solution on YOUR website!
Not really enough information given. The two real zeros can be a parabola or quadratic function
and the simplest function can simply use those two zeros for the binomial roots.

The example using the two imaginary zeros will not cross nor touch the x-axis, and because no
information is given to say anything about the leading coefficient, you do not have information
about vertex as max or as min. Pick either and as simple as possible.
y=%28x-2i%29%28x%2B2i%29
y=x%5E2-%282i%29%5E2
y=x%5E2-4i%5E2
y=x%5E2-4%2A%28-1%29
highlight%28y=x%5E2%2B4%29

This as picked, opens upward, HAS NO REAL ZEROS, therefore has no x-axis intercepts, and has
a vertex minimum at (0,4). Same as the graph of y=x^2 but raised upward by 4 units.

graph%28300%2C300%2C-8%2C8%2C-2%2C14%2Cx%5E2%2B4%29