The equations are
y = -[tex]x^{2}[/tex] - 1
y = 2[tex]x^{2}[/tex] -4
The graphs of the solutions (x, y) of these equations are 2 parabolas, since the right hand side expressions are polynomials of degree 2.
The solution/s of the system are the x-coordinates of the point/s of intersection of the parabolas.
The solutions of the first equation form a parabola looking downwards (since the coefficient of x^2 is -), and the second, a parabola opening upwards (since the coefficient of x^2 is +).
We can draw both parabolas, but to find the solution we still need to solve the system algebraically.
The algebraic solution of the system is:
-[tex]x^{2}[/tex] -1 =2[tex]x^{2}[/tex] - 4
3[tex]x^{2}[/tex] -3 =0
3([tex]x^{2}[/tex] -1) = 0
[tex]x^{2}[/tex] -1 =0 , so
the solutions are x=-1 and x=1.
If we are allowed to use a graphic calculator, we can draw both graphs and point at the solution.
