Creating Quadratic Equations from Graphs

In the last lesson, we presented the graphs depicted below. Remember that for graphs a and c, the horizontal axis represents fire severity, The vertical axis represents tree density and tree height, respectively

Green removed .png

Today, you will delve into applying function transformations and various mathematical techniques to construct models for graphs a) and c). Subsequently, you'll develop a model for graph d) and apply it to problem-solving tasks.

Creating Linear and Quadratic Models and Use Models to Solve Problems: Example #1

Writing Line Equations
Find the y-intercept
Find the slope
Real World Model: Domain

In Module 2, you reviewed methods for writing the equation of a line.

For example, if you are given two points on a line, you can find the slope of the line as follows:

Given the points (1,8) and (3,5), the slope of the line is given by:

$LaTeX: m=\frac{8-5}{1-3}=\frac{3}{-2}$

The equation of the line containing these points is

$LaTeX: y=-\frac{3}{2}x+b$

where b is the y-intercept.

You can find the y-intercept by using one of the two given points on the line and substituting the coordinates of that point into the equation of the line. For example, if you use the point (1,8), you substitute 1 in for x and 8 in for y and you have:

$LaTeX: 8=-\frac{3}{2}\left(1\right)+b$

$LaTeX: 8=-\frac{3}{2}+b$

$LaTeX: 8+\frac{3}{2}=b$

$LaTeX: 9\frac{1}{2}=b$

The equation of the line is

$LaTeX: y=-\frac{3}{2}x+9\frac{1}{2}$

The line is shown in the graph below.

Screenshot 2024-05-16 at 8.32.13 AM.png

To find a model for the linear function in Graph c) below, you will estimate two points as follows:

One point is the y-intercept (0,255) and the other is the x-intercept (105,0).

Using these two points, you can find the slope of the line as follows:

$LaTeX: m=\frac{255-0}{0-105}=\frac{255}{-105}\approx-2.43$

Since you estimated the y-intercept to be 255, you have the equation of the line.

The graph of the line is shown below:

Screenshot 2024-05-16 at 8.37.38 AM.png

You will limit the domain of your model to match the real-world context. Since the fire severity cannot be less than zero or more than 100, you limit the domain of the model. Recall that the domain is the set of values for x.

The new domain for your real-world model is:

$LaTeX: 0\le x\le100$

Below is a graph of the model with its limited domain:

Screenshot 2024-05-16 at 8.37.38 AM.png

Now you can use your model to answer the following questions:

Predict the tree height if the fire severity is 70%.

Evaluating the model at 70%, we have $LaTeX: f\left(70\right)=-2.43\left(70\right)+255=85$ cm

For what percentage of fire severity corresponds to the tree height 100cm?

To answer this question, we will need to substitute 100 in for y and solve for x, as follows:

LaTeX: 100=-2.43x+255

LaTeX: 100-255=-2.43x

LaTeX: -155=-2.43x

$LaTeX: \frac{-155}{-2.43}=x$

$LaTeX: x\approx63.79$

or

$LaTeX: 63.8\%$

x is approximately equal to 63.79

Creating Linear and Quadratic Models and Use Models to Solve Problems: Example #2

Vertex Form of Quadratic Functions
Find the value of a
Real World Model: Domain

In the graph below, you see the Tree Density in stems/hectare as a function of the Fire Severity in percent.

It appears to be quadratic. To find a model for this curve, you will use your knowledge of quadratic functions.

Screenshot 2024-05-16 at 11.24.18 AM.png

You recall that the vertex form of a quadratic function is

$LaTeX: f\left(x\right)=a\left(x-h\right)^2+k$

The vertex of the parabola is

$LaTeX: \left(h,k\right)$

Looking at the graph you see that the vertex of this graph is approximately (0,6000).

A quadratic model for this graph is given by

$LaTeX: f\left(x\right)=a\left(x-0\right)^2+6000$

or

$LaTeX: f\left(x\right)=ax^2+6000$

Now you will estimate another point on the graph and substitute that point into the model to find the value of a.

You will use the point (50,2000). Substituting that point into your model, you have:

$LaTeX: 20,000=a\left(50\right)^2+6000$

Solving for a, we have:

$LaTeX: 20,000=a\left(50\right)^2+6000$

$LaTeX: 20,000-6000=a\left(50\right)^2$

$LaTeX: 14,000=a\left(50\right)^2$

LaTeX: 14,000=2500a

$LaTeX: \frac{14,000}{2500}=a$

LaTeX: 5.6=a

Your model is

$LaTeX: f\left(x\right)=5.6x^2+6000$

A graph of our model is shown in the graph below.

Screenshot 2024-05-16 at 11.35.59 AM.png

As you did in the first example, you will limit the domain of your model to match the real-world context. Since the fire severity cannot be less than zero or more than 100, you limit the domain of the model. Recall that the domain is the set of values for x. The new domain for your real-world model is

$LaTeX: 0\le x\le100$

Below is a graph of the model with the limited real-world domain.

Screenshot 2024-05-16 at 11.35.59 AM.png

Now you can use your model to answer the following questions:

What is the tree density if the fire severity is 50%?

Evaluating the model at 50%, we have

$LaTeX: f\left(50\right)=5.6\left(50\right)^2+6000=20,000$ stems per hectare.

Self-Check:

The Bakery Problem: The Fresh Bakery currently sells 1, 000 huge chocolate chip cookies each week for $0.50 each. The bakery would like to increase its revenue from the cookie sales by increasing this price. A survey convinced the manager that for every $0.10 increase, the bakery would sell 70 fewer cookies each week. The graph of the function for the revenue from the sale of huge chocolate cookies is shown in the graph below.

Screenshot 2024-10-04 at 9.19.24 PM.png

Problem A:

Using your knowledge of quadratic functions write a model for the revenue as a function of x. where x is the number of 10 cent increases per cookie. Use technology to check your answer. Round each number in your equation to three decimal places. Recall the vertex form of a quadratic is given by:

$LaTeX: f\left(x\right)=a\left(x-h\right)^2+k$

Solution A:

Explanation: From the graph we see the vertex is (4.643,650.893). Use the vertex form of the function we have

$LaTeX: f\left(x\right)=a\left(x-4.643\right)^2+650.893$

Using the point (0,500), we can find a by substituting the point into the equation.

$LaTeX: 500=a\left(0-4.643\right)^2+650.893$

$LaTeX: 500=a\left(-4.643\right)^2+650.893$

$LaTeX: 500-650.983=a\left(21.557\right)$

LaTeX: -150.983=a

$LaTeX: -150.983=a\left(21.557\right)$

$LaTeX: -\frac{150.983}{21.577}=a$

LaTeX: -6.997=a

Problem B:

Use your model to find the answer to the three questions below:

Question 2:

1. At what price should the cookies be sold to maximize the revenue from the sale of huge chocolate cookies for the bakery? Explain your reasoning.

Solution 1:

Since the vertex or maximum occurs at (4.643,650.893), the number of 10 cent increases in the price of a cookie is 4.643 at the maximum revenue. So the price of a cookie would be 50 cents plus 4.643 times 10 cents. That means the price of the cookie should be approximately 96 cents in order to maximize revenue.

Question 2:

2). What is the maximum revenue?

Solution 2:

The maximum revenue is $650.89 since that is the y-value of the maximum point or the vertex.

Question 3:

3). What is the revenue if they raise the price of each cookie to $1.20? Notice this represents seven 10 cent increases in the price (0.50+0.70=$1.20). Explain your reasoning.

Solution 3:

$272.17

Since $1.20 represents seven 10-cent increases. We substitute 7 into the function:

$LaTeX: f\left(7\right)=-6.997\left(7-4.643\right)^2+650.89=-6.997\left(2.357\right)^2+650.89$

$LaTeX: -6.997\left(2.357\right)^2+650.89=-38.87+650.89$

LaTeX: -38.87+650.89=612.02

When you have completed the above lesson, click the next button in Canvas to progress.