y<x^2+8x+20

y>=-x^2-10x-20
what is the solution of the system of inequalities?
Then says to graph but not sure if the graph is shaded or not.

I have a ti-84 and don't know how to use it.

Got it

To find the solution of the system of inequalities y < x^2 + 8x + 20 and y >= -x^2 - 10x - 20, we can start by graphing the two equations separately.

First, let's graph y = x^2 + 8x + 20. This is a quadratic equation, and its graph will be a parabola. To find the x-intercepts, we set y = 0 and solve for x:

0 = x^2 + 8x + 20

Using the quadratic formula, x = (-b ± √(b^2 - 4ac))/(2a), where a = 1, b = 8, and c = 20:

x = (-8 ± √(8^2 - 4(1)(20)))/(2(1))
x = (-8 ± √(64 - 80))/2
x = (-8 ± √(-16))/2

Since the discriminant is negative, there are no real solutions for x. This means that the parabola does not cross the x-axis.

Next, let's graph y = -x^2 - 10x - 20. Again, this is a quadratic equation, and its graph will also be a parabola. To find the x-intercepts, we set y = 0 and solve for x:

0 = -x^2 - 10x - 20

Using the quadratic formula, x = (-b ± √(b^2 - 4ac))/(2a), where a = -1, b = -10, and c = -20:

x = (-(-10) ± √((-10)^2 - 4(-1)(-20)))/(2(-1))
x = (10 ± √(100 - 80))/(-2)
x = (10 ± √(20))/(-2)
x = (10 ± 2√(5))/(-2)
x = -5 ± √(5)

So, the x-intercepts of this parabola are x = -5 + √(5) and x = -5 - √(5).

Now, let's plot these graphs on a coordinate plane:

The graph of y = x^2 + 8x + 20 will be a parabola that does not cross the x-axis.

The graph of y = -x^2 - 10x - 20 will be a parabola that crosses the x-axis at x = -5 + √(5) and x = -5 - √(5).

To shade the area that satisfies both inequalities, we need to consider the overlapping region of the two graphs. Since y < x^2 + 8x + 20 defines an area below the parabola and y >= -x^2 - 10x - 20 defines an area on or above the parabola, the shaded region will be the area on or above the parabola y = -x^2 - 10x - 20.

Thus, the solution to the system of inequalities is the region on or above the parabola y = -x^2 - 10x - 20.