Section 4.4 solutions
1) First, let me graph the function, and describe the features of the graph.
This graph tracks the flight of the ball. The x-coordinate of each point represents a time in seconds, and
the y-coordinate of each point the height of the ball. The point (1,64) represents the ball being 64 feet
high after 1 second. Although the graph goes on forever, only the region displayed is relevant to the
problem. Points beneath the x-axis would tell you when the ball is underground, as this doesn’t happen,
those points may be ignored.
a) How long does it take for the ball to hit the ground.
Answer: 5 seconds. (the x-coordinate of the x-intercepts are the times when the ball as on the ground.
Algebra to find the answer – set the equation equal to 0
0 = -16t2 + 80t (now factor out GCF of -16t)
0 = -16t(t – 5)
-16t = 0
t=0
t–5=0
t = 5 (the ball is at ground level at 0 seconds and 5 seconds)
b) When does the ball reach its maximum height? Answer: 2.5 seconds.
The ball reaches its maximum height at the x-coordinate of the vertex. I could graph and use the
maximum feature on my calculator.
−𝑏
Algebra to find x-coordinate of vertex – use formula x-coordinate of vertex = 2𝑎 in this case a = -16 and
b = 80
−80
x-coordinate of vertex = 2(−16)= 2.5
c) what is the maximum height of the ball? The maximum height occurs at the vertex.
Answer: 100 feet
The maximum height is the y-coordinate of the vertex. I could get this by using the maximum feature of
my calculator, or I can plug in the x-coordinate of the vertex into the function
Maximum height = -16(2.5)2 + 80(2.5) = 100
3) This problem is essentially the same problem as problem 1. I will solve using algebra, however, a
graph can be drawn to get the answers.
a) how long will it take the ball to reach a height of 35 meters?
Replace h with 35 and solve for t
35 = -4.9t2 +30t + 1.6
-35
-35
2
0 = -4.9t + 30t -33.4
I will now use the quadratic formula with a = -4.9, b = 30,
𝑡=
−30±√302 −4(−4.9)(−33.4)
2∗−4.9
𝑡=
−30±√245.36
−9.8
𝑡=
−30±15.66
−9.8
𝑡=
−30+15.66
−9.8
𝑜𝑟
−30−15.66
−9.8
t = 1.46 or t = 4.66
Answer: 1.46 seconds and 4.66 seconds
b) How long will it take to land?
Find the non-zero x-intercept to do this
0 = -4.9t2 + 30t + 1.6 (I will use the quadratic formula again)
𝑡=
−30±√302 −4(−4.9)(1.6)
2∗−4.9
𝑡=
−30±√931.36
−9.8
𝑡=
−30±30.52
−9.8
c = -33.4
𝑡=
−30+30.52
−9.8
𝑜𝑟
−30−30.52
−9.8
t= -.05 or 6.18
Answer: it takes 6.18 seconds to return to the ground
(note the first answer t = -.05 does not apply as it represents a time before the ball was thrown, also the
1.6 in the equation tells me this ball started at 1.6 meters off the ground, rather than starting at the
ground.)
c) determine the maximum height the object reaches:
This is the y-coordinate of the vertex.
−𝑏
I will find the x-coordinate of the vertex using the formula 2𝑎
x-coordinate of vertex
−30
2∗−4.9
= 3.06
now plug the 3.06 into the equation to get the y-coordinate of the vertex.
h= -4.9(3.06)2 + 30(3.06) + 1.6
h = 47.52
answer: maximum height 47.52 meters
5) This just wants me to find the positive x-intercept. I do this by setting the function equal to 0.
0 = -16t2 + 64t + 80
I will factor 0 = -16(t2 – 4t - 5)
0 = -16(t+1)(t-5)
t+ 1 = 0 or t – 5 = 0
t = -1 or t = 5
answer: it will hit the ground in 5 seconds
7) This also wants me to find the positive x-intercept.
0 = -16t2 + 160
0 = -16(t2 – 10)
-16 = 0 has no solution
t2 – 10 = 0 solve with square roots
t2 = 10
t = ±√10
only the positive answer makes sense for this problem
answer t = √𝟏𝟎 seconds, about 3.16 seconds
9) Each point on the graph of the function is of the form
(number of units sold in a week, profit from selling that many units)
The graph of the function is a parabola, and the vertex of the parabola is a maximum point. The solution
we are looking for will be contained in the vertex.
x-coordinate of vertex – number of units to sell in a week to maximize profit
y-coordinate of vertex – maximum profit
I will use the
−𝑏
2𝑎
formula to find the x-coordinate of the vertex where a = -1 and b = 50
x-coordinate of vertex =
−50
2(−1)
= 25
answer to part a: 25 units will maximize profit
y-coordinate of vertex -(25)2 +50(25) = 625
answer to part b: maximum profit will be $625
11) The function is a parabola, the vertex is a minimum point. Each point on the parabola has the form
( production level, cost at that production level)
Hence the vertex will contain the solutions I need.
−𝑏
x-coordinate of vertex can be computed using the 2𝑎 formula and it will give the answer to the question
how many units should be produced.
x-coordinate of vertex =
−(−10)
=20
2∗.25
answer to part a: produce 20 units to minimize cost
y-coordinate of vertes C(20) = 0.25(20)2 -10(20) + 800 = 700
answer to part b: minimum cost is $700
13)
a) scatterplot
13b) linear regression line is y = 3x+36 (with r of ..016)
quadratic regression equation is y = -19.29x2 +118.71x -99 (with r2 of .96)
13c) the quadratic regression will be a better fit, the data doesn’t appear to be linear and the low r
value confirms that
13d)
I’m surprised how poorly the function fits the data.
13e) the y-coordinate of the vertex of the parabola is the maximum height. I will be a bit lazy and use
my calculator to find this. I will graph the function y = -19.29x2 + 118.71x – 99 and use the maximum
feature from my calculator.
Answer: 83.63 feet is the maximum height according to the formula, of course I know this isn’t correct
as the data tells me the ball goes higher.
15a)
15b) linear regression equation y = -1.22x + 56.20 *r = -.03)
Quadratic regression equation y = -56.21x2 +139.31x +9.35
15c) A parabola looks like it will fit the data better and the quadratic regression equation will be the one
to use.
15d)
15e) Plug in 1.75 for x into the equation to get your answer.
y = -56.21(1.75)2 + 139.31(1.75) +9.35
y = 81.00
answer: 81 mg/l