anonymous
  • anonymous
I need help with this question, The graph below shows the value of Edna's profits f(t), in dollars, after t months: graph of quadratic function f of t having x intercepts at 6, 0 and 18, 0, vertex at 12, negative 36, and passes through point 21, 41.25 What is the closest approximate average rate of change for Edna's profits from the 18th month to the 21st month?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
whees the graph??
anonymous
  • anonymous
I don't know how to put it on here but the part that says graph of quadratic function f of x that is what is on the graph.
anonymous
  • anonymous
click on attach file i can't understand u

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anonymous
  • anonymous
Oh ok sorry on my computer the screen is small so i didn't see that okay one sec and i will put the graph.
anonymous
  • anonymous
ok :3
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
Here is the graph
anonymous
  • anonymous
ok wait up
anonymous
  • anonymous
ok
anonymous
  • anonymous
rate of change= slope
anonymous
  • anonymous
do you know how to find the slope?
anonymous
  • anonymous
since you are not answering i will show ya
anonymous
  • anonymous
slope=\[\frac{ y2-y1 }{ x2-x1 }\]
anonymous
  • anonymous
so your two points from 18 months to 21 months is: (0,18) and (21,41.25).
anonymous
  • anonymous
Hey sorry it wouldn't let me type lol
anonymous
  • anonymous
plug the points into the slope formula and you get:\[\frac{ 41.25-18 }{ 21-0 } =\frac{ 23.25 }{ 21 }\]
anonymous
  • anonymous
thats the rate o change :3
anonymous
  • anonymous
g'night btw
anonymous
  • anonymous
Oh okay thank you! Lol
anonymous
  • anonymous
:)

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