Question

What is the focus of the following graph?

Answer 1

@jim_thompson5910 only two more =)

Answer 2

This is another parabola. The vertex is (0,0), so (h,k) = (0,0) giving us h = 0 and k = 0


So we can say

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

y = a(x-0)^2 + 0

y = ax^2


We know that the point (2,-1) lies on the graph, so x = 2 and y = -1. Plug these values in


y = ax^2

-1 = a(2)^2

-1 = 4a

-1/4 = a

a = -1/4


So a = -1/4, which means that the equation is \[\Large y = -\frac{1}{4}x^2\]

Answer 3

Oh misread, one sec

Answer 4

Now convert to the form 4p(y-k) = (x-h)^2


y = -(1/4)x^2


(y - 0) = -(1/4)(x - 0)^2


-4(y - 0) = (x - 0)^2


4*(-1)(y - 0) = (x - 0)^2


So p = -1, which tells us that we go down one unit from the vertex (along the axis of symmetry) to get to the focus. So subtract 1 from the y coordinate of the vertex (0) to get 0 - 1 = -1


So the focus of the parabola is (0,-1)

Answer 5

Thank you very much!! =)

Answer 6

you're welcome