Help please with this question
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The equations are
,
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:
, so
the solutions are x=-1 and x=1.
The graph of the system is drawn using desmos.com
If we are allowed to use a graphic calculator, we can draw both graphs and point at the solution.
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:
the solutions are x=-1 and x=1.
The graph of the system is drawn using desmos.com
If we are allowed to use a graphic calculator, we can draw both graphs and point at the solution.