# Solve for the system of inequalities graphically: (i) y < x^2 (ii) 9x^2-36x+4y^2-16y < -16show domain and range

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To solve the system of inequalities graphically, you graph each relation and shade the appropriate area creating the solution set. Then the points that lie in both solution sets are all solutions to the system.

(1) The graph of `y<x^2` : First graph `y=x^2` with a dotted line. This is a parabola opening up, vertex at the origin. Choosing a point not on the graph, e.g. (4,0), we find that all points on the same side of the parabola (the "outside") are solutions. (`0<4^2` is true, so all points in that half-plane are solutions. Had we chosen (0,3) we would find that `3<0` was false, and thus no points in that half-plane are solutions.)

(2) Graph the relation `9x^2-36x+4y^2-16y=-16` with a dotted line.

Recognize this as a conic section. Use completing the square to write in standard form:

`9x^2-36x+4y^2-16y=-16`

`9(x^2-4x)+4(y^2-4y)=-16`

`9(x^2-4x+4)+4(y^2-4y+4)=-16+36+16`

`9(x-2)^2+4(y-2)^2=36`

`((x-2)^2)/4+((y-2)^2)/9=1`

This is an ellipse; the major axis is parallel to the y-axis with a=3; the minor axis parallel to the x-axis with b=2, centered at (2,2). The inequality indicates that we should shade inside the ellipse: trying the point (2,2) yields `9(4)-36(2)+4(4)-16(2)<-16` as a true statement.

**(3) Thus the solution set is the set of points inside of the ellipse, but "outside" the parabola.**

Using a graphing utility, we find that the parabola intersects the ellipse at `x~~2.23,y~~4.98` and `x~~.41,y~~.17`

The domain for the solution set is approximately `.41433777<x<4` while the range is approximately `-1<y<4.979826`

Note: be careful when using a graphing utility to graph the inequalities if you must enter them as functions; you would have `sqrt((36-9(x-2)^2)/4)+2<1` and `-sqrt((36-9(x-2)^2)/4)+2>1`