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 > = -x^2 - 2 y >…

Question

asked Dec 6, 2021 13.3k views

1 Answer

Andref · Answer 1 · 2021-12-11T10:21:30+0000

Answer:

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

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

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

Explanation:

A vertex form equation of a parabola is of the form

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

with vertex at (h,k).

If a parabola opens downwards, then a<0.

If the vertex of such parabola is below the x-axis, then it has no real solution, because it will not intersect the x-axis.

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

has vertex at (-4,-2), which is below x-axis.

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

has vertex at (-4,2)--->above x-axis

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

vertex at (4,-2) ---> below x-axis

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

(0,-2)----> below x-axis