Quadratic function apq question?

Question

A quadratic function passes through the point (3, -28) and has a vertex at (6, -64). Write the function in the form y = a(x - p)2 + q. What are the values of a, p and q?

Answer 1

y = a(x-p)^2 + q

this in standard form where vertex(p, q)

(p,q) = (6,-64)

so p = 6

q = -64

y = a(x-6)^2 -64

this passes through point(3, -28)

-28 = a(3-6)^2 - 64

-28 = 9a - 64

9a = 36

a = 36/9

= 4

Answer 2

(x-h)^2 = 2p(y-k)^2 where (h,k) is vertex and p = distance between focus and directrix.

(x-6)^2 = 2p(y+64)
When y = -28, x = 3 so,
(3-6)^2 =2p(-28+64)
9 = 2p(36) --> 2p = 1/8
So (x-6)^2 = (1/8) (y+64), or
y = 8(x-6)^2 -64
So the vertex is (6,-64) which is your p and q
and a = 8 = distance between focus and directrix so focus is at (6,60) and directrix is the line y = -68. Thevertex is halfway between the focus and the directrix on the axis of symmetry.

Answer 3

p = 6
q = -64

-28 = a(3-6)^2-64
36 = a(-3)^2
a = 4

y = 4(x-6)^2 - 64

Answer 4

use the formula

y2-y1 = y-y1
-------- ------
x2-x1 x-x1

u have two point assume 1 point as (x1, y1) and the other point as (x2, y2)
and get the equation