CPhill

As as alternative way to do this....note that the distance between the vertex and directrix = 5 units = "p"

Since the directrix is above the vertex....the coordinates of the focus =

( x coordinate of the vertex, y coordinate of the vertex - p )  =

(0 , 0 - 5 ) =

(0, - 5)

The parabola opens downward because the directrix is above the vertex.....to find the focus we have

( x coordinate of the vertex,  y coordinate of the vertex - y value of the directrix ) = 

(0, 0 - the y value of the directrix)  =

(0 , 0 - 5 )  =

(0 , - 5)  =   the focus

This is always confusing....

Note that  we have

(x + 7)^2 + 4

The form that we are really going for is

( x - h)^2 + k        where ( h, k) is the vertex

So....

(x - (-7) )^2 + 4     produces the vertex (h, k)  =  (-7, 4)

But this is just the same as

(x + 7)^2 + 4

Remember that a " + "  moves the x coordinate of the vertex to the left  and a " - "  moves the x coordinate of the vertex to the right

To get a feel for this....look at the graph here :  https://www.desmos.com/calculator/4ryl8slhym

y = ( x + 2)^2  shifts the vertex of y = x^2 to the left 2 units

y = ( x - 2)^2  shifts the vertex of y = x^2 to the right by 2 units

y = x^2    has the vertex at (0, 0)

So....we have

y = (x - 0)^2 + 0

This graph translates the vertex to the left 7 units and up 4 units

So...we have

y = ( x + 0 + 7) ^2 + ( 0 + 4 ) 

y = ( x + 7)^2 + 4

No prob....I didn't invent this junk....I just try to wade my way through it, too!!!

No big deal....I'm sure Rom has worked harder than that at some point in his existence.....

We have the form

y = cos (Bx)

B =  2pi / period

Note that the period = pi

So

B =  2pi / pi = 2

So....the correct answer is 

y = cos (2x)

LOL!!!...cruel to make Rom work that hard !!!

Focus

Directrix

I always think that the "mathematicians" devise stuff like this to make math seem more "dark and deep" than it needs to be....also.....it sells a lot of math textbooks.....

x^3 + 2x^2 - 7x - 2

We have something called the Rational Roots Theorem here that might help us....without going into detail ...the possible  rational roots are   1, - 1 , 2 or   -2

I see that 1 is not a root

I see that -1  isn't either

But 2 is a root because  (2)^3 + 2(2)^2  -7(2) - 2  =  8 + 8 - 14 - 2   = 0

So....using synthetic division, we can find  the remaining polynomial

2   [    1    2    - 7    -2 ]

                2     8     2

______________________

         1     4     1      0

The remaining polynomial is   x^2 + 4x + 1

We can complete the square on x to find the other two roots

x^2 + 4x + 4 =  - 1 + 4

(x + 2)^2 =  3    take both roots

x + 2 = ±√ 3      subtract 2 from both sides and we get the other two solutions

x = √3  - 2     or   x  = -√3   -  2