Question

Which of the following are possible equations of a parabola that has no real solutions and opens downward?
y>= -(x+4)^2 - 2
y>= -(x+4)^2 + 2
y>= -x2-2
y>=(x-4)^2 + 2
y>= -(x-4)^2 - 2

Answer 1

Answer:

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

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

Step-by-step explanation:

The vertex form of a parabola is given as

$y = a {(x - h)}^{2} + k$

If this parabola opens downward, then a<0.

This means for a function opening downward not to have a solution, the k value must also be negative and taking a square root of a negative number will not give real solution.

From the given options, examples of such functions are:

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

and

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

Answer 2

Answer:it’s y>=-(x+4)^2-2

Step-by-step explanation:

I checked it in my graphing calculator and it works. I’m also doing the program sooo