Math
posted by Wendy
1. If the point (3,30) is on the graph y=(x1)^2+14, where will the point be on y=(x8)^26? I don't understand what this is asking and how I can find this.
2. A rental business charges $12 per canoe and averages 36 rentals a day. For every 50cent increase in rental price, it will lose two rentals a day. What price would yield the maximun revenue? So I have a feeling this has something to do with quadratic equations, and it's max, but I don't know how to incorporate this data into the vertex formula.
3. You have a 1200foot roll of fencing and a large field. You want to make two paddocks by splitting a rectangular enclosure in half. What are the dimensions of the largest such enclosure? I tried drawing a diagram and using the area to no avail.

Reiny
The way I read it ....
y = (x8)^2  6 is the result if the parabola
y = (x1)^2 + 14 has been moved 7 units to the right and 20 units down, so the point (3,30) went along for the ride, so
(3,30) > (4,10)
check:
old vertex = (1,14)
new vertex = (8,6) > 7 to the right, 20 down
or
sub (4,10) into new equation
LS = 10
RS = (48)^2  6
= 166 = 10
= LS
#2
let the number of 50 cent increases in the rental be x
new cost per canoe = 12 + .5x
new number rented = 50  2x
profit = (12 + .5x)(502x)
= 600  24x + 25x  x^2
P = x^2 + x + 600
There is your quadratic function, I assume you know how to find the vertex, and once you have your vertex you can pick off your answer.
#3.
Let the length of the whole field be y, let each of the widths be x
so 2y + 3x = 1200 > y = (12003x)/2 = 600  (3/2)x
area = A = xy
A = x(600  (3/2)x )
expand to get the quadratic function again.
Notice you will have a negative x^2 term, so it opens down, giving you a maximum of A 
Wendy
For #2, I found the vertex (1/2, 600.25) but my answer key says $10.50. How?

Wendy
also, why is the new # rented 502x? Wouldn't it be 362x or something like that?

Wendy
and for #3, what does the 2y and 3x represent? I got the answer right thank you, but for future reference I don't know how I can find the equation..
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