Question

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

Answer 1

We have to identify the parabola that:

a) has no real solutions (it has complex roots).

b) opens downward.

The equations are in vertex form, so to have no real solutions and be downward, the vertex should have a value of y that is less than 0:

If the vertex had a value of y greater than 0, if it opens downward, there should be two values of x so that f(x) = 0. Then, this would be the real solutions.

The vertex form can be written as:

`y=a(x-h)^2+k`

where (h,k) are the coordinates of the vertex.

Then, we need k <= 0.

The equations that satisfy this are:

y = -(x + 4)² - 2

y = -x² - 2

y = -(x - 4)² - 2

The second condition is that the parabola opens downward. This means a quadratic coefficient "a" with negative value.

All this three equations satisfy this condition.

We can graph them and check if the conditions are satisfied:

Answer:

There are 3 equations that satisfy the two conditions:

y = -(x + 4)² - 2

y = -x² - 2

y = -(x - 4)² - 2