Please explain. I want to understand the steps.
1. A frog jumps from a rock to the shore of a pond. Its path is given by the equation y=(-5/72)x^2+(5/3)x, where x is the horizontal distance in inches, and y is the height in inches. What is the frog's maximum height? How far had it traveled horizontally when it reached its maximum height?
2. A local bakery determines that when the price of a package of cookies is $12, it sells on average 220 packages of cookies per day. For every $0.50 increase in the price, it sells on average 10 packages of cookies less. Determine the price that maximizes the bakery's profit. Determine the maximum profit.
3.Find the quadratic function with the vertex (7,10) and with point (10,28) on its graph.
looks like you are studying quadratic function.
You should have learned a method to find the vertex of such a funtion
here is a quick way
for y = ax^2 + bx + c
the x of the vertex is -b/(2a)
so in your case
x = (-5/3)/(-5/32) = 32/3
sub back in to get
y = 80/9
y is your height,
x is your horizontal distance
2. let the number of $0.50 increases be n
profit = (12 + .5n)(220 - 10n)
expand to get a quadratic, repeat the method I showed you in #1
3. let the function be
y = a(x-7)^2 + 10 , in the usual vertex form
but (10,28) lies on it
so 28 = a(10-7)^2 + 10
solve for a, and you got it
For 2 I did the function and got n=-1. then i substitute it in and got 11.5 is that the answer? if so to which of the 2 questions?
You are right, n = -1
Since I defined n as the number of .50 "increases" we have to interpret that as a "decrease" of 50 cents.
you can verify this by making a chart
cost - number - profit
12.00 220 ---- 2640
12.50 210 ---- 2625
11.50 230 ---- 2645
11.00 240 ---- 2640
etc
so the price should be $11.50, selling 230 of them for a maximum profit of $2645
btw, (11.5 , 2645) would be the vertex of your function
Sure! Let's go through each question step by step.
1. To find the frog's maximum height and the distance it traveled horizontally when it reached that height, we can use the vertex form of a quadratic equation. Start with the given equation: y = (-5/72)x^2 + (5/3)x.
The vertex form for a quadratic equation is y = a(x - h)^2 + k, where (h, k) represents the vertex. In this case, we are given that the vertex is (h, k) = (7, 10).
To convert the given equation to vertex form, we need to complete the square. Let's do that:
y = (-5/72)x^2 + (5/3)x
= (-5/72)(x^2 - (5/12)x)
Now, to complete the square, we need to find the value that needs to be added and subtracted inside the parentheses to make it a perfect square trinomial. To do that, we take half of the coefficient of the x-term, square it, and add/subtract it inside the parentheses:
y = (-5/72)(x^2 - (5/12)x + (5/24)^2 - (5/24)^2)
= (-5/72)((x - 5/24)^2 - 25/576)
= (-5/72)(x - 5/24)^2 + 25/1152
Now we can see that the equation is in vertex form: y = a(x - h)^2 + k, where (h, k) = (5/24, 25/1152).
From the equation, we can see that the vertex represents the maximum point of the parabola. Therefore, the frog's maximum height is k = 25/1152 inches.
To find the distance traveled horizontally when it reached its maximum height, we look at the x-value of the vertex, which is h = 5/24. So the frog traveled 5/24 inches horizontally when it reached its maximum height.
2. To determine the price that maximizes the bakery's profit and the maximum profit, we can use a similar method.
Let's assume that the price of a package of cookies is P dollars. And let's assume that the number of packages sold per day is S.
From the given information, we can establish a relationship between the price and the number of packages sold:
S = 220 - 10(P - 12)/0.50
Simplifying this equation, we have:
S = 220 - 20(P - 12)
Now, we need to find the price that maximizes the bakery's profit. The profit is given by the equation:
Profit = Price * Sales
Since we have the number of packages sold (S) as a function of the price (P), we can write the profit equation as:
Profit = P * (220 - 20(P - 12))
Expanding and simplifying this equation, we have:
Profit = 220P - 20P^2 + 240P - 2640
Combining like terms, we get:
Profit = -20P^2 + 460P - 2640
To find the price that maximizes the profit, we need to find the vertex of this quadratic equation. The x-coordinate of the vertex, in this case, represents the price. We can use the formula x = -b / 2a to find it.
In our equation, a = -20 and b = 460, therefore:
P = -460 / (2 * -20)
P = 11.5
Therefore, the price that maximizes the bakery's profit is $11.50.
To find the maximum profit, we substitute this price back into the profit equation:
Profit = -20(11.5)^2 + 460(11.5) - 2640
Profit = $4,290
Therefore, the maximum profit the bakery can make is $4,290.
3. To find the quadratic function with the given vertex and point on its graph, we can use the vertex form of a quadratic equation.
The vertex form of a quadratic equation is y = a(x - h)^2 + k, where (h, k) represents the vertex. In this case, we are given that the vertex is (h, k) = (7, 10).
Substituting the vertex values into the vertex form, we get:
y = a(x - 7)^2 + 10
Now, we are given that another point on the graph is (10, 28). We can substitute this point's coordinates into the equation to find the value of a.
28 = a(10 - 7)^2 + 10
28 = a(3)^2 + 10
28 = 9a + 10
9a = 28 - 10
9a = 18
a = 2
Now that we have the value of a, we can substitute it back into the vertex form equation:
y = 2(x - 7)^2 + 10
Therefore, the quadratic function with the given vertex (7, 10) and with the point (10, 28) on its graph is y = 2(x - 7)^2 + 10.