Can we illustrate this pair of inequalities on a graph?

Can we illustrate this pair of inequalities on a graph?

Question

Describe and indicate by shading, on the usual axes, the set of points $S_1$ defined by $(x-2)(y+3) \geq 0$. $S_2$ is the set of points defined by $x^2+ y^2 \leq 25$. Find the range of values of $p$ if the point $(3, p) \in S_1 \cap S_2$.

[Editorial Note: Here by $(3, p) \in S_1 \cap S_2$, we mean that the point $(3,p)$ is contained in both $S_1$ and $S_2$.]

Suggestion

What might the boundary of $S_1$ look like? If we know the boundary of a region, trying a few examples can often help us to see which side of the boundary is the required region.

$S_2$ is the set of points defined by $x^2 + y^2 \leq 25$. Find the range of values of $p$ if the point $(3, p) \in S_1 \cap S_2$.

Sketching the region $S_2$ on the same axes may well help here.

Solution

The boundary of the region is given by $(x-2)(y+3)=0$, so either $x=2$ or $y=-3$. So $S_2$ is bounded by two straight lines.

We have $(x-2)(y+3) \geq 0$ when either $x-2\geq 0$ and $y+3 \geq 0$, or $x-2\leq 0$ and $y+3 \leq 0$. Alternatively, we could pick a point in each of the four regions formed by the two straight lines to determine whether or not it is in $S_1$. For example, $(0,0)$ has $(0-2)(0+3)=-2<0$, so the origin does not lie in $S_1$.

Using either approach, we deduce that $S_1$ is the shaded region in this diagram:

The boundary of the region (the lines $x = 2$ and $y = -3$) is included in $S_1$.

$S_2$ is the set of points defined by $x^2 + y^2 \leq 25$. Find the range of values of $p$ if the point $(3, p) \in S_1 \cap S_2$.

The set $S_2$ is the interior and boundary of a circle centred on the origin with radius $5$.

Thus the intersection of $S_1$ and $S_2$ (the points in both $S_1$ and $S_2$) is the area of overlap:

Thus the range of values of $p$ if the point $(3, p) \in S_1 \cap S_2$ is given by the $y$-coordinates of the points in the intersection of the overlap region with the line $x = 3$:

It’s easy to see that the line $x=3$ leaves the overlap region at $(3,4)$ (since this is on the circle) and at $(3,-3)$ (which is on the line $y=-3$), and both of these points are included, so $-3 \leq p \leq 4$.