Question

quadratic applications - vertical motion

Answer 1

I need to find the domain and range of R(p)= -2p^2+200p

Answer 2

What class are you taking?

Answer 3

Algebra 2 Pre-AP

Answer 4

ok, recall, the general form for a quadratic

y(x) = ax^2 + bx + c

a= -2
b = 200

Answer 5

What do you know how to do so far?

Answer 6

I know how to use the quadratic formula

Answer 7

but like since there's no c im not sure what to do.....

Answer 8

Since you are asked to find the domain and the range, if you find the vertex of the parabola, and knowing that a=-2 (negative number), so the thing opens down,
then you can give the range.
The domain is gonna be all real numbers x = -infinity to x=infinity

Answer 9

Do you remember how to find where the vertex point is?

Answer 10

Yeah don't you use a formula like -b/2a or something like that

Answer 11

right, the Axis of symmetry will have an X value as what you said, the vertex lies on the axis of symmetry

Answer 12

ok and the vertex is the range right? actually idk

Answer 13

well, it will be a limit of the range.

In the general form y(x) = ax^2 + bx + c
\[if ~~a<0,~~then ~opens ~down\]\[if~a>0,~then~opens~upwards\]

Answer 14

ok so it will open down

Answer 15

Using the formula you gave for finding the vertex, you get
vertex (h,k) = (50, 5000)

and a is less than zero, so it opens down

Answer 16

In this case when it opens downwards, the vertex point will be the maximum value for the range.

Answer 17

5000 is the largest value for y, and it can be any number less than that. So the range is
\[(-\infty ,~5000]\]

Answer 18

ok and that means the domain is 50?

Answer 19

would it be ok if i wrote the range like [0, 5000]

Answer 20

and for some reason i thought at first the range was [0, 200] idk why

Answer 21

no, the range can be any number less than or equal to 5000.

\[y \le 5000\]

Answer 22

ohhhh ok i see hmm

Answer 23

THe vertex is at (50, 5000) and the parabola opens down. So the range is any number less than or equal to 5000.

Answer 24

The function will never give you values greater than 5000,
\[R(p) \le 5000\]

Answer 25

this is kind of like what im doing, just so you know

Answer 26

except my question is about money and has nothing with velocity, which threw me off

Answer 27

Yeah, they are modeled the same way. Projectile motion is a quadratic function.

Answer 28

The domain of a general quadratic function like r(p) = -2p^2+200p
is all real numbers.
However, the situation given in the problem may limit the domain..

Like in projectile motion, the time will be larger than zero, so the domain will be not all real numbers, but numbers greater than or equal to zero.
It depends on the problem given.

Answer 29

So that's kind of why i was thinking the range was [0, 200] because the maximum sell revenue using the function would be 200? but yeah im not sure

Answer 30

and then usually you use the quadratic formula to get the domain but there is no "c"

Answer 31

What are the variables? revenue r as a function of price p?

Answer 32

The quadratic formula will give you the zeros, or the X-axis intercepts of the parabola, which may be needed in some problems

Answer 33

Ok let me type the question: A manufacturer of tricycles finds that the weekly revenue is a function of the price of the tricycle, p. The equation for the revenue is R(p)= -2p^2+200p

Answer 34

so i have to find the domain and range of this and also find the maximum sales revenue which i assume is also the range

Answer 35

Ok, in this problem , the domain and range will be limited because of a couple things.

The domain (possible values for Price p), will be larger than zero. You can't have a negative price.

The range (resulting values for Revenue R), will also be larger than zero. You dont get a negative revenue.

Answer 36

So here you will indeed need to find the X-intercepts of the fuction using the quadratic formula or factoring.

Answer 37

i would like to use the quadratic formula but im not sure how since there is no c

Answer 38

ok, The function is
R(p) = -2p^2 + 200p

This would be easier to find the zeroes from factoring first.

R(p) = -2p * (p - 100)

You want to find when R(p) = 0

0 = -2p *( p - 100 )

Answer 39

If you have something like
0 = A * B

That will be true if either A=0, or, B = 0

Answer 40

\[0 = -2p*(p-100)\]\[-2p = 0~~~~or~~~~(p-100)=0\]
\[p = 0~~~~~or~~~~~p=100\]

Answer 41

So you have the vertex, and the p intercepts.

(h,k) = (50 , 5000)

p = 0 or p = 100

Answer 42

The graph Is this...

Answer 43

Ahhhh ok i think i get it....

Answer 44

So since the domain is 100 does that mean the maximum sales revenue is 100?

Answer 45

Using that function for revenue. The possible Prices p are\[0 \le p \le 100\]
and the resulting range is
\[0 \le R(p) \le 5000\]

Answer 46

The maximum revenue occurs at the Vertex point, and is 5000 dollars, when the price p is 50 dollars.

Answer 47

Not sure if you have ever seen this site, but there are a bunch of vids that can help you out...
https://www.khanacademy.org/math/algebra/algebra-functions/domain_and_range/v/domain-and-range-of-a-function-given-a-formula

Answer 48

https://www.khanacademy.org/math/algebra/quadratics/solving_graphing_quadratics/v/finding-the-vertex-of-a-parabola-example

Answer 49

Thank you so much, I will let you know if i get a good grade, wish me luck ha

Answer 50

haha, good luck. If you have more questions, you can add @DanJS to them and Fan me if you want