SOLUTION: The profit on a watch is given by P = x2 – 13x – 80 and where x is the number of watches sold per day. How many watches were sold on a day when there was a $50 loss?
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Question 188303: The profit on a watch is given by P = x2 – 13x – 80 and where x is the number of watches sold per day. How many watches were sold on a day when there was a $50 loss?
Found 2 solutions by jojo14344, Mathtut:
Answer by jojo14344(1513) (Show Source): You can put this solution on YOUR website!
Let known P=Y=Profit or Losses
Solving for "x" for number of watches. More importantly to see the X-Intercepts.
Given:
By Quadratic, where
Discriminant: b^2-4ac= -13^2 - 4(1)(-80)=169+320=489
Then,
X-Intercepts = > (17.55,0) & (-4.55,0)
For Y-Intercept:
Let fx=0 ---->
Y-Intercept = > (0,-80)
We now get the vertex thru Vertex-Form, , where
Going back to our eqn:
, complete the square, by taking "Half" of the middle constant then "squared":
In our eqn, the middle constant is 13, taking half of it = , then squared it= . We "add" and also "subtract" to the eqn:
--->follows
In our eqn, we have vertex, h=x=6.5 & k=y=-122.25
We see graph,
The Y axis, is where the Profit or Losses.
So if you have a loss of $50, that equals to -50 on the Y axis.
Make a straigth line from -50 on the Y axis to the right until it touches the curve. As it touched the curve, extend the line downwards until it touches the X axis.
This will show the number of watches for the loss sale of $50.00.
----> 15 watches sold for a loss of $50.00
Let's check,
Go back our Eqn, & substitute x=15:
, Loss
Thank you,
Jojo
Answer by Mathtut(3670) (Show Source): You can put this solution on YOUR website!
:
:
:
so x=15 and -2....since you cant sell a negative number of watches
:
x=15 watches
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