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Smoke Jumpers parachute into locations to
suppress forest fires
When they exit the airplane, they are in free
fall until their parachutes open
https://www.youtube.com/watch?v=sovMpY2
QH-I
If the jumper exits the airplane at a height of
554m, how long will the jumper be in free fall
before the parachute opens at 300m?
1. Determine the quadratic relation to model
the height, H, of the smoke jumper at time t
2. Determine the length of time that the
jumper is in free fall.
Notice: the problem contains information
about the vertex: he jumps at a height of 554m
at time t = 0.
Note Also: a = -0.5g = -9.8m/s2 (acceleration
due to gravity)
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H
H
H
H
H
H
=
=
=
=
=
=
a(t – h)2 + k
a(t – 0)2 + 554
at2 + 554
-0.5(9.8)t2 + 554
-0.5(9.8)t2 + 554
-4.9t2 + 554
Therefore, H = -4.9t2 + 554 is an equation in
standard form that models this relationship.
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Recall, we’re looking for the length of time
that the jumper is in free fall.
The parachute opened at 300m, so substitute
300 for H and then find t.
H = -4.9t2 + 554
300 = -4.9t2 + 554
300 -554 = -4.9t2 + 554 -554
-254 = -4.9t2
−254
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−𝟒.𝟗
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=
−4.9𝑡 2
−𝟒.𝟗
So the jumper is in free
fall for about 7.2
seconds.
t2 = 51.84
t2= 51.84
t = 7.2
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We used 0 for the tcoordinate because we
started counting when he
jumped out of the plane
We were only considering
the right-half of this graph
◦ t≠0
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A coffee shop sells a special blend of coffee
for $2.60 per mug. The shop sells about 200
mugs per day. Customer surveys show that
for every $0.05 decrease in the price, the
shop will sell 10 more mugs per day.
a) Determine the maximum daily revenue
from coffee sales and the price per mug for
this revenue.
b) Write an equation in both standard form
and vertex form to model this problem. Then
sketch the graph.
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Let x represent the number of $0.05
decreases in price
Revenue = (price)(mugs sold)
Price: 2.60 – 0.05x
Mugs Sold: 200 + 10x
Revenue = (2.60 – 0.05x)(200 + 10x)
We want to find the maximum revenue so we
need to vertex
We can start by finding the zeros, and then
finding the middle of the zeros to determine
the equation of the axis of symmetry
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We find the zeros (x-intercept) by setting y=0
Revenue = (2.60 – 0.05x)(200 + 10x)
0 = (2.60 – 0.05x)(200 + 10x)
So either 2.60 – 0.05x = 0 or 200 + 10x = 0
If 2.60 – 0.05x = 0, x = 52
If 200 + 10x = 0, x = -20
The vertex is in the middle of the zeros, so:
𝑥=
52+(−20)
2
= 16
We know the vertex is at x = 16, the r value is:
 Revenue = (2.60 – 0.05x)(200 + 10x)
r = (2.60 – 0.05(16))(200 + 10(16))
r = (1.80)(360)
r = 648
a)
Therefore, the maximum daily revenue is $648.
The price per mug to maximize revenue is:
2.60 – 0.05x at x = 16, so:
2.60 – 0.05(16) = 1.80
So the coffee shop should sell each mug of coffee
for $1.80 to achieve the maximum revenue of $648.
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b) Write an equation in both standard form and vertex
form to model this problem. Then sketch the graph.
Here, the vertex is (16, 648) = (h, k), so we can write:
r = a(x – 16)2 + 648
We know that when x = 0, r = (2.60 – 0.05x)(200 + 10x),
so:
r = (2.60)(200) = 520, so a point is (0, 520) – use to find
“a”
r = a(x – 16)2 + 648
520 = a(0 – 16)2 + 648
520 = a(– 16)2 + 648
-128 = 256a
-0.5 = a
So, the equation in vertex form is r = -0.5(x – 16)2 + 648
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To get standard form, expand the equation in vertex form.
r = -0.5(x – 16)2 + 648
r = -0.5(x2 – 32x + 256) + 648
r = -0.5x2 + 16x - 128 + 648
r = -0.5x2 + 16x + 520 is the equation in standard form.
We can sketch the graph because we know the vertex: (16, 648),
x-intercepts (x = 52 & x = -20) and the y-intercept (0, 520):
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All quadratic relations can be expressed in
vertex form and standard form
Quadratic relations that have zeros can also
be expressed in factored form
For any parabola, the value of a is the same in
all three forms of the equation of the
quadratic relation