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.
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)
old vertex = (1,14)
new vertex = (8,-6) ---> 7 to the right, 20 down
sub (4,10) into new equation
LS = 10
RS = (4-8)^2 - 6
= 16-6 = 10
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
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.
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
For #2, I found the vertex- (1/2, 600.25) but my answer key says $10.50. How?
also, why is the new # rented 50-2x? Wouldn't it be 36-2x or something like that?
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..