anonymous
  • anonymous
FAN AND MEDAL The graph below shows a company's profit f(x), in dollars, depending on the price of pens x, in dollars, being sold by the company: graph of quadratic function f of x having x intercepts at ordered pairs 0, 0 and 6, 0. The vertex is at 3, 120 Part A: What do the x-intercepts and maximum value of the graph represent? What are the intervals where the function is increasing and decreasing, and what do they represent about the sale and profit? Part B: What is an approximate average rate of change of the graph from x = 3 to x = 5, and what does this rate represent?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
anonymous
  • anonymous
@triciaal yayy!!
anonymous
  • anonymous
do you know how to do this?

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More answers

triciaal
  • triciaal
the x-intercept means y = 0 f(x) = 0 f(x) = profit maximum value = highest y value = highest profit
anonymous
  • anonymous
so x-intercept means theres 0 profit?
triciaal
  • triciaal
vertex is at 3, 120 so when x = 3 y = 120 3 pens maximum profit $120
triciaal
  • triciaal
yes
anonymous
  • anonymous
ok and what about the increasing intervals?
triciaal
  • triciaal
question is asking how does y change as x changes
anonymous
  • anonymous
the more pens sold the higher the profit?
anonymous
  • anonymous
but aren't they asking for the intervals where the function is increasing and decreasing
triciaal
  • triciaal
up to the limit of 3 maximum so x 0 to 3 increase
triciaal
  • triciaal
x more than 3 profit decrease
anonymous
  • anonymous
x more than 3 profit decrease ?
triciaal
  • triciaal
look at the graph
anonymous
  • anonymous
ohh ok
anonymous
  • anonymous
now part B
anonymous
  • anonymous
wait but what does the sale an profit represent when it increases and decreases
triciaal
  • triciaal
change in y divided by change in x f(5) - f(3) divided by (5-3)
anonymous
  • anonymous
but what does the sale an profit represent when it increases and decreases
anonymous
  • anonymous
nvm that
anonymous
  • anonymous
but last question
anonymous
  • anonymous
what is the equation for the graph?
anonymous
  • anonymous
@triciaal ?
triciaal
  • triciaal
yes you tell me
anonymous
  • anonymous
i don't know how to make an equation for it
triciaal
  • triciaal
|dw:1439405938151:dw|
anonymous
  • anonymous
ohh ok one second
triciaal
  • triciaal
|dw:1439405998517:dw|
anonymous
  • anonymous
f(x)=(x-3)^2+120?
triciaal
  • triciaal
|dw:1439406035909:dw|
anonymous
  • anonymous
how do i find a?
anonymous
  • anonymous
@triciaal how do i find a??
anonymous
  • anonymous
@ganeshie8 can u plz help
anonymous
  • anonymous
@welshfella
anonymous
  • anonymous
all i need is the equation and i don't know how to get to it @welshfella
welshfella
  • welshfella
equation for the graph can be written as f(x) = a(x - 3)^2 + 120 because the vertex is at (3,120) and a is a constant to be found
anonymous
  • anonymous
yes i did that too but i don't know ow to find a
triciaal
  • triciaal
|dw:1439406519835:dw|
anonymous
  • anonymous
f(x)=(x-0)(x-6)=0 is the equation?
anonymous
  • anonymous
?
anonymous
  • anonymous
hello
welshfella
  • welshfella
the zeroes of the equation are x=0 and x = 6 so we can plug these into f(x) = a(x -3)^2 + 120 0 = (-3)^2 a + 120
anonymous
  • anonymous
i don't need the zeroes though, i only need the equation itself and thats it
anonymous
  • anonymous
f(x)=(x-3)^2+120 is this the equation
welshfella
  • welshfella
well if you find the value of a you'll get the equation
welshfella
  • welshfella
no - you need to find a
anonymous
  • anonymous
AHH whats a?!?!?
welshfella
  • welshfella
0 = 9 a + 120 a = ?
anonymous
  • anonymous
-13.33?
welshfella
  • welshfella
f(x) = a(x -3)^2 + 120
welshfella
  • welshfella
yea so we have f(x) -13.333(x - 3)^2 + 120
anonymous
  • anonymous
ok thank you so much!!!!
welshfella
  • welshfella
yw
anonymous
  • anonymous
@welshfella can u give @triciaal a medal for me
welshfella
  • welshfella
ok
anonymous
  • anonymous
thanks! @welshfella
triciaal
  • triciaal
you are welcome

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