# Math

posted by Wendy

1. If the point (-3,30) is on the graph y=(x-1)^2+14, where will the point be on y=(x-8)^2-6? 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 50-cent 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 1200-foot 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.

1. Reiny

The way I read it ....
y = (x-8)^2 - 6 is the result if the parabola
y = (x-1)^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 = (4-8)^2 - 6
= 16-6 = 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)(50-2x)
= 600 - 24x + 25x - x^2

P = -x^2 + x + 600

#3.
Let the length of the whole field be y, let each of the widths be x
so 2y + 3x = 1200 ---> y = (1200-3x)/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

2. Wendy

For #2, I found the vertex- (1/2, 600.25) but my answer key says \$10.50. How?

3. Wendy

also, why is the new # rented 50-2x? Wouldn't it be 36-2x or something like that?

4. 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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