Ihave 3 questions here , one of them i just need my answer checked on and the other two i really need with.

Problem 1) 

The narrowest graph happens when the leading coefficient is furthest from 0. In this case, that happens to be 4 which is the coefficient in choice B. 

See the attached image "figure1" for the graph. The functions are color coded

y = -x^2 is in red

y = 4x^2 is in blue

y = (1/4)x^2 is in green (note: 1/4 = 0.25; so y = (1/4)x^2 is the same as y = 0.25x^2)

y = (1/9)x^2 is in purple (note: 1/9 = 0.11 approximately; so y = (1/9)x^2 is roughly the same as y = 0.11x^2)

As you can see in figure1, the blue graph corresponding to y = 4x^2 is the most narrowest. 

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Problem 2) 

See figure2 for the graph. The image is also attached along with figure1.

Figure2 shows a parabola that opens upward and the lowest point is at (0,2) which is the vertex. This point is shown in red as point A.

y = x^2+2 is the same as y = 1(x-0)^2 + 2. That second equation is in the form y = a(x-h)^2 + k where a = 1, h = 0, k = 2

So the vertex is (h,k) = (0,2). The fact that 'a' is positive (a = 1) indicates that the parabola opens upward and this implies the vertex is the lowest point.

Therefore the vertex (0,2) is the minimum

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Problem 3) 

You have the correct answer here. Nice work. 

Where are the following equations?