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)
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
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 = (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..
1. To find where the point (-3, 30) will be on the graph y = (x - 8)^2 - 6, you can follow these steps:
Step 1: Start with the equation y = (x - 8)^2 - 6.
Step 2: We know that (-3, 30) is a point on the graph, so substitute -3 for x and 30 for y in the equation.
30 = (-3 - 8)^2 - 6
Step 3: Simplify the equation:
30 = (-11)^2 - 6
30 = 121 - 6
30 = 115
Step 4: Since 30 does not equal 115, this means that the point (-3, 30) is not on the graph of y = (x - 8)^2 - 6. Therefore, there is no corresponding point on this graph.
2. To find the price that would yield the maximum revenue for the rental business, you can follow these steps:
Step 1: Let's assume the initial rental price is $12 and the number of rentals is 36.
Step 2: The revenue is calculated by multiplying the rental price by the number of rentals: Revenue = Price x Quantity.
Step 3: Now, for every 50-cent increase in rental price, the business loses 2 rentals. This means that for every additional 50 cents added to the price, the number of rentals will decrease by 2.
Step 4: We can form an equation for the revenue in terms of the rental price. Let's call the rental price "x" in dollars:
Revenue = (36 - 2(x - 12))(x)
The first part, 36 - 2(x - 12), represents the adjusted number of rentals based on the price increase. We subtract 12 from x to reflect the number of 50-cent increases from the initial price of $12.
Step 5: Simplify the equation:
Revenue = (36 - 2x + 24)x
= (-2x + 60)x
= -2x^2 + 60x
Step 6: This equation is in quadratic form, with the coefficient of x^2 being -2. To find the maximum revenue, we can use the vertex formula. The x-coordinate of the vertex is given by x = -b / (2a), where a = -2 and b = 60.
x = -60 / (2(-2))
= -60 / (-4)
= 15
Step 7: So, the rental price that would yield the maximum revenue is $15.