Exercise 11 Page 642

We want to solve the given system of inequalities by graphing. Let's graph each of inequality one at a time.

Inequality (I)

We can write the boundary curve for Inequality (I) by replacing the less than or equal to sign with an equals sign. The above equation is an equation of an ellipse. Let's write it in a standard form. The denominator of the $y\text{-}$variable is greater than the denominator of the $x\text{-}$variable. Therefore, we are given the equation of the vertical ellipse. Let's find necessary information to draw a graph.

Equation of the Vertical Ellipse	$\frac{(y-k{)}^2}{a^2}+\frac{(x-h{)}^2}{b^2}=1,$ $a$ and $b$ positive, with $a>b$	$\frac{(y-0{)}^2}{4^2}+\frac{(x-0{)}^2}{2^2}=1$
Center	$(h,k)$	$(0,0)$
Length of Major Axis	$2a$ units	$2(4)=8$ units
Length of Minor Axis	$2b$ units	$2(2)=4$ units
Foci	$(h,k\pm c),$ $c^2=a^2-b^2$	$(0,0\pm c),$ $c^2=4^2-2^2$

Let's calculate the value of $c$ and find the foci. Since we will add and subtract $c$ to find the foci, we kept the principal root when calculating its value. We will now use the information we found to draw the ellipse. Because we have a non-strict inequality, the boundary curve will be solid. Now that we have the boundary curve, we need to determine which region to shade. To do so, we will use $(\text{-}1,1)$ as a test point. If the point satisfies the inequality, we will shade the region that contains the point. If not, we will shade the opposite region. Since the substitution produced a true statement, we will shade the region that contains the point $(\text{-}1,1).$

Inequality (II)

Notice, that Inequality (II) is a quadratic inequality. We can write the boundary curve for Inequality (II) by replacing the greater than or equal to sign with an equals sign. Then, we can identify the values of $a,$ $b,$ and $c.$ Knowing that $a=\text{-}1,$ $b=0,$ and $c=2,$ we can find the vertex. To do so, we will need to think of $y$ as a function of $x,$ $y=f(x).$ Let's substitute the values of $a$ and $b$ in the formula for the $x\text{-}$coordinate of the vertex. The $x\text{-}$coordinate of the vertex is $0.$ Now, let's find the $y\text{-}$coordinate by substituting $0$ for $x$ into the quadratic equation for the boundary line. The vertex is $(0,2).$ With this information, we know that the axis of symmetry of the parabola is the vertical line $x=0.$ Next, let's find two more points on the curve, one on each side of the axis of symmetry.

$x$	$\text{-}{x}^2+2$	$y=\text{-}{x}^2+2$
$\text{-}2$	$\text{-}(\text{-}2{)}^2+2$	$\text{-}2$
$2$	$\text{-}{2}^2+2$	$\text{-}2$

The points $(\text{-}2,6)$ and $(2,6)$ are on the parabola. Let's plot the points and connect them with a smooth curve. Since we have a non-strict inequality, the curve will be solid.

Once again, we will check which region we should shade. To do so, we will use $(0,0)$ as a test point. Since the substitution did not produce a true statement, we will shade the region which does not contain the point $(0,0).$

Solution

The solution set is the overlapping region.