From Förberedande kurs i matematik 1

The inequality \displaystyle y\le \text{1}-x^{\text{2}} defines the area under and on the curve \displaystyle y=\text{1}-x^{\text{2}}, which is a parabola with a maximum at \displaystyle \left( 0 \right.,\left. 1 \right). We can rewrite the other inequality \displaystyle x\ge \text{ 2}y-\text{3} as \displaystyle y\le \text{ }{x}/{2}\;+{3}/{2}\; and it defines the area under and on the straight line \displaystyle y=\text{ }{x}/{2}\;+{3}/{2}\;.

Of the figures above, it seems that the region associated with the parabola lies completely under the line \displaystyle y=\text{ }{x}/{2}\;+{3}/{2}\; and this means that the area under the parabola satisfies both inequalities.

NOTE: If you feel unsure about whether the parabola really does lie under the line, i.e. that it just happens to look as though it does, we can investigate if the \displaystyle y -values on the line \displaystyle y_{\text{line}}=\text{ }{x}/{2}\;+{3}/{2}\; is always larger than the corresponding \displaystyle y -value on the parabola \displaystyle y_{\text{parabola}}=\text{1}-x^{\text{2}} by studying the difference between them:

\displaystyle y_{\text{line}}-y_{\text{parabola}}=\text{ }\frac{x}{2}+\frac{3}{2}-\left( \text{1}-x^{\text{2}} \right)

If this difference is positive regardless of how \displaystyle x is chosen, then we know that the line's \displaystyle y -value is always greater than the parabola's \displaystyle y -value. After a little simplification and completing the square, we have

\displaystyle \begin{align} & y_{\text{line}}-y_{\text{parabola}}=\text{ }\frac{x}{2}+\frac{3}{2}-\left( \text{1}-x^{\text{2}} \right)=x^{2}+\frac{1}{2}x+\frac{1}{2} \\ & =\left( x+\frac{1}{4} \right)^{2}-\left( \frac{1}{4} \right)^{2}+\frac{1}{2}=\left( x+\frac{1}{4} \right)^{2}+\frac{7}{16} \\ \end{align}

and this expression is always positive because \displaystyle \frac{7}{16} is a positive number and \displaystyle \left( x+\frac{1}{4} \right)^{2} is a quadratic which is never negative. In other words, the parabola is completely under the line.