Answer: The correct options are (1) (5,10), (2) (3,-3), (3) x = -1, (4)
, (5) 21s and (6) 0, -1, and 5.
Explanation:
Te standard form of the parabola is,
.....(1)
Where, (h,k) is the vertex of the parabola.
(1)
The given equation is,

Comparing this equation with equation (1),we get,
and 
Therefore, the vertex of the graph is (5,10) and the fourth option is correct.
(2)
The given equation is,


To make perfect square add
, i.e.,
. Since there is factor 3 outside the parentheses, so subtract three times of 9.


Comparing this equation with equation (1),we get,
and 
Therefore, the vertex of the graph is (3,-3) and the fourth option is correct.
(3)
The given equation is


To make perfect square add
, i.e.,
. Since there is factor 4 outside the parentheses, so subtract three times of 1.


Comparing this equation with equation (1),we get,
and 
The vertex of the equation is (-1,3) so the axis is x=-1 and the correct option is 2.
(4)
The given equation is,

To make perfect square add
, i.e.,
.



Therefore, the correct option is 4.
(5)
The given equation is,

It can be written as,

It is a downward parabola. so the maximum height of the function is on its vertex.
The x coordinate of the vertex is,



Therefore, after 21 seconds the projectile reach its maximum height and the correct option is first.
(6)
The given equation is,


Use factoring method to find the factors of the equation.



Equate each factor equal to 0.

Therefore, the zeros of the given equation is 0, -1, 5 and the correct option is 2.