Question

Examine the following system of inequalities.

Answer 1

Answer:

Dotted linear inequality shaded below passes through (0, 4) & (4,3). Dotted parabolic inequality shaded above passes through points (negative 6,4), (negative 4, 0) & (negative 2, 4).

Step-by-step explanation:

Hello! Let me help you to find the correct option to this problem. First of all, we have the following system of inequalities:

$\left\{ \begin{array}{c}y< -\frac{1}{4}x+4\\y>(x+4)^{2}\end{array}\right.$

To solve this, let's write the following equations:

FIRST:

$y=-\frac{1}{4}x+4$

This is a linear function written in slope-intercept form as $y=mx+b$. So, the slope $m=-\frac{1}{4}$ and the y-intercept is $b=4$. Since in the inequality we have the symbol < then the graph of the line must be dotted. To get the shaded region, let's take a point, say, $(0, 0)$ and let's test whether the region is above or below the graph. So:

$y< -\frac{1}{4}x+4 \\ \\ Let \ x=y=0 \\ \\ 0$

Since the expression is true, then the region is the one including point $(0, 0)$, that is, it's shaded below.

SECOND:

$y=(x+4)^{2}$

This is a parabola that opens upward and whose vertex is $(-4,0)$. Since in the inequality we have the symbol > then the graph of the parabola must be dotted. Let's take the same point $(0, 0)$ to test whether the region is above or below the graph. So:

$y>(x+4)^{2} \\ \\ Let \ x=y=0 \\ \\ 0>(0+4)^2\\ \\ 0>16 \ False!$

Since the expression is false, then the region is the one that doesn't include point $(0, 0)$, that is, it's shaded above

____________________

On the other hand, testing points (0, 4)  and (4,3) on the linear function:

$y=-\frac{1}{4}x+4 \\ \\ \\ \bullet \ (0,4): \\ \\ y=-\frac{1}{4}(0)+4 \therefore y=4 \\ \\ \\ \bullet \ (4,3): \\ \\ y=-\frac{1}{4}(4)+4 \therefore y=3$

So the line passes through these two points.

Now, testing points (negative 6,4), (negative 4, 0) & (negative 2, 4) on the parabola:

$y=(x+4)^2 \\ \\ \\ \bullet \ (-6,4): \\ \\ y=(-6+4)^2 \therefore y=(-2)^2 \therefore y=4 \\ \\ \\ \bullet \ (-4,0): \\ \\ y=(-4+4)^2 \therefore y=0 \\ \\ \\ \bullet \ (-2,4): \\ \\ y=(-2+4)^2 \therefore y=(2)^2 \therefore y=4$

So the line passes through these three points.

Finally, the shaded region is shown below.