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).
Hello! Let me help you to find the correct option to this problem. First of all, we have the following system of inequalities:
[tex]\left\{ \begin{array}{c}y< -\frac{1}{4}x+4\\y>(x+4)^{2}\end{array}\right.[/tex]
To solve this, let's write the following equations:
[tex]y=-\frac{1}{4}x+4[/tex]
This is a linear function written in slope-intercept form as [tex]y=mx+b[/tex]. So, the slope [tex]m=-\frac{1}{4}[/tex] and the y-intercept is [tex]b=4[/tex]. 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, [tex](0, 0)[/tex] and let's test whether the region is above or below the graph. So:
[tex]y< -\frac{1}{4}x+4 \\ \\ Let \ x=y=0 \\ \\ 0<-\frac{1}{4}(0)+4 \\ \\ 0<4 \ True![/tex]
Since the expression is true, then the region is the one including point [tex](0, 0)[/tex], that is, it's shaded below.
[tex]y=(x+4)^{2}[/tex]
This is a parabola that opens upward and whose vertex is [tex](-4,0)[/tex]. Since in the inequality we have the symbol > then the graph of the parabola must be dotted. Let's take the same point [tex](0, 0)[/tex] to test whether the region is above or below the graph. So:
[tex]y>(x+4)^{2} \\ \\ Let \ x=y=0 \\ \\ 0>(0+4)^2\\ \\ 0>16 \ False![/tex]
Since the expression is false, then the region is the one that doesn't include point [tex](0, 0)[/tex], that is, it's shaded above
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[tex]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[/tex]
So the line passes through these two points.
[tex]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[/tex]
So the line passes through these three points.
Finally, the shaded region is shown below.
