Question

Determine the equation of the parabola in the form y= a(x-p)^2 +pq

a) the parabola has a vertex of (0,-3) and a x-intercept of 4

b) the x-intercept are x = -4 and x=2 and the graph is congruent to y= -2/3(x+1)^2 -3

Answer 1

Please check for typo in the original equation:
is it
y= a(x-p)^2 +pq
or
y= a(x-p)^2 +q

Answer 2

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

Answer 3

Since the vertex form of the parabola is
y=a(x-h)^2+k where (h,k) is the vertex, we conclude that for
y=a(x-p)^2+q with vertex at (0,-3), then
p=0, q=-3, so
y=ax^2-3
Since one zero is at x=4, and the zeroes of a parabola are symmetrical about the vertex, we conclude that there are two zeroes, x=4, and x=-4, resulting in the equation
y=a(x-4)(x+4) = a(0)^2-3
Solving for a,
a=-3/(-16)=3/16
=>
y=(3/16)x^2-3

b.
y=-2/3(x+1)^2 -3 is already in the vertex form,
a=-2/3
p=-1
q=-3

Answer 4

thank u sooo much! :D

Answer 5

You're welcome! :)