# Question cb310

Aug 22, 2016

For problems of this type, you need to first graph, and then shade the solution region using test points.

The parabola $y = {x}^{2} - 8$ has a vertex at $\left(0 , - 8\right)$. Using a table of values, you will be able to graph it very precisely. The table of values may look like the following:

While you graph, make sure that you have a dotted line, since this inequality has the symbol ">" and not "≥".

For the second equation, x + y ≤ 1, I recommend solving for $y$ so that you convert to slope intercept form; it's easier to graph this way.

Isolating y:

y ≤ 1 - x

You can now place the y intercept, which is $1$, and start placing points using the slope of the line (Vertical change: -1, horizontal change: 1).

You can then connect these dots using a complete line, not a dotted one, since the symbol used in this case is ≤ and not $<$.

As for the shading, usually for quadratic inequalities I would select test points and for linear inequalities I would shade through logical induction. In y ≤ 1- x, we would shade below the line, because we want all points where $y$ is smaller than $1 - x$.

For $y > {x}^{2} - 8$, select test points, one inside the parabola and one on the outside.

Let Test Point 1 be (0, 0)

0 >^? 0^2 - 8

$0 > - 8$

So, we shade inside the parabola.

Once this is done, the solution to the inequality is the region where the two shaded areas overlap.

Here is what you should have in the way of graphing, except your graph should have the parabola as a dotted line (my program can for some reason not draw this properly).

As for points of intersection, I think the graph says it all. A few examples are $\left(0 , - 6\right)$ and $\left(- 2 , - 2\right)$.

Hopefully this helps!

Aug 23, 2016

$x \in \left(\frac{- 1 - \sqrt{37}}{2} , \frac{- 1 + \sqrt{37}}{2}\right) , y \in \left[{x}^{2} - 8 , 1 - x\right]$

#### Explanation:

Another way, algebraically:

$x + y \le 1$

$\implies y \le 1 - x$

$\implies 1 - x \ge y > {x}^{2} - 8 \text{ (*)}$

$\implies 1 - x > {x}^{2} - 8$

$\implies {x}^{2} + x - 9 < 0$

The roots of ${x}^{2} + x - 9 = 0$ are $x = \frac{- 1 \pm \sqrt{37}}{2}$. As ${x}^{2} + x - 9 > 0$ when the magnitude of $x$ grows large (in the positive or negative direction), then it must be the case that ${x}^{2} + x - 9 < 0$ only between the two roots above:

$\frac{- 1 - \sqrt{37}}{2} < x < \frac{- 1 + \sqrt{37}}{2}$

Given a restriction on $x$, we now can add a restriction on $y$ in terms of $x$. However, in our previous work, we already found such a restriction at $\text{(*)}$. Thus, the set of points satisfying both inequalities is given by

$x \in \left(\frac{- 1 - \sqrt{37}}{2} , \frac{- 1 + \sqrt{37}}{2}\right) , y \in \left[{x}^{2} - 8 , 1 - x\right]$

Note that the restrictions on the $x$ and $y$ values cannot be completely independent of each other. For example, if we solved for the upper and lower bounds on $y$, we would find that

$- \frac{33}{4} \le y < \frac{3 + \sqrt{37}}{2}$,

But points such as $\left(1 , 2\right)$ are not in the solution set, even though there are points in the solution set with the $x$-coordinate $1$ and points with the $y$-coordinate $2$.