Standard form is, hold a sec
x=2 is directix
that means it opens left or right
so we must use
(y-k)²=4p(x-h)
where vertex is (h,k) and p is distance from focus to vertex
also shortest distance from vertex to directix
the shortest distance from focus to directix is 2p
if p>0 then the parabola opens right
if p<0 then pareabola opens left
so
(-2,0) and x=2
the distance is 4
4/2=2
p=2
wait, positive or negative
focus is to the left of the directix so p is negative
p=-2
vertex is 2 to the right of the focus and 2 to the left of directix
vertex is (0,0)
so
(y-0)²=4(-2)(x-0) or
y²=-8x is da equation
not sure what form is standard tho
it takes 5 seconds to reach a height of 40 ft. and It takes 2.5 seconds to reach maximum height.
Let h(t) represent the height of the ball at time t.
Given that the height is given by:
h(t) = -16t² + 80t + 40
1) For the ball to reach a height 0f 40 ft, h(t) = 40, hence:
40 = -16t² + 80t + 40
16t² - 80t = 0
t(t - 5) = 0
t = 0 or t = 5
Hence it takes 5 seconds to reach a height of 40 ft.
2) The maximum height is at h'(t) = 0,
h'(t) = -32t + 80
-32t + 80 = 0
32t = 80
t = 2.5
It takes 2.5 seconds to reach maximum height.
Find out more at: brainly.com/question/11535666
F = 1.8F + 32
F - 1.8F = 1.8F - 1.8F + 32
-0.8F = 32
-0.8F / -0.8= 32 / -0.8
F = -40
