Look At This Table:$\[ \begin{tabular}{|c|c|} \hline $x$ & $y$ \\ \hline -7 & -294 \\ \hline -6 & -216 \\ \hline -5 & -150 \\ \hline -4 & -96 \\ \hline -3 & -54 \\ \hline \end{tabular} \\]Write A Linear \[$(y=mx+b)\$\], Quadratic

Feb 26, 2025 by ADMIN 230 views

Exploring Linear and Quadratic Relationships in a Given Table

In mathematics, relationships between variables are often expressed through equations. These equations can be linear or quadratic, depending on the nature of the relationship. A linear equation is a polynomial equation of degree one, while a quadratic equation is a polynomial equation of degree two. In this article, we will explore the linear and quadratic relationships in a given table and discuss the methods used to determine these relationships.

A linear relationship is a relationship between two variables where one variable is a constant multiple of the other variable. In other words, if we have two variables x and y, a linear relationship between them can be expressed as y = mx + b, where m is the slope of the line and b is the y-intercept.

To determine the linear relationship in the given table, we can use the method of finding the slope and y-intercept. The slope of a line is a measure of how much the line rises (or falls) vertically over a given horizontal distance. It can be calculated using the formula:

m = (y2 - y1) / (x2 - x1)

where (x1, y1) and (x2, y2) are two points on the line.

Calculating the Slope

Using the given table, we can calculate the slope of the line by choosing two points, say (-7, -294) and (-6, -216). The slope can be calculated as follows:

m = (-216 - (-294)) / (-6 - (-7)) = (-216 + 294) / (-6 + 7) = 78 / 1 = 78

Calculating the Y-Intercept

Now that we have the slope, we can calculate the y-intercept by substituting the slope and one of the points into the equation y = mx + b. Let's use the point (-7, -294).

-294 = 78(-7) + b -294 = -546 + b b = 252

Linear Equation

Now that we have the slope and y-intercept, we can write the linear equation as follows:

y = 78x + 252

This equation represents the linear relationship between the variables x and y in the given table.

A quadratic relationship is a relationship between two variables where one variable is a polynomial of degree two in the other variable. In other words, if we have two variables x and y, a quadratic relationship between them can be expressed as y = ax^2 + bx + c, where a, b, and c are constants.

To determine the quadratic relationship in the given table, we can use the method of finding the coefficients a, b, and c. The coefficients can be found by substituting the values of x and y from the table into the equation y = ax^2 + bx + c and solving for a, b, and c.

Finding the Coefficients

Using the given table, we can substitute the values of x and y into the equation y = ax^2 + bx + c and solve for a, b, and c. Let's use the points (-7, -294), (-6, -216), and (-5, -150).

-294 = a(-7)^2 + b(-7) + c -294 = 49a - 7b + c

-216 = a(-6)^2 + b(-6) + c -216 = 36a - 6b + c

-150 = a(-5)^2 + b(-5) + c -150 = 25a - 5b + c

Solving the System of Equations

We now have a system of three equations with three unknowns. We can solve this system using substitution or elimination. Let's use the elimination method.

First, we can multiply the first equation by 2 and the second equation by 3 to make the coefficients of c equal.

-588 = 98a - 14b + 2c

-648 = 108a - 18b + 3c

Now, we can subtract the first equation from the second equation to eliminate c.

-60 = 10a - 4b

Next, we can multiply the first equation by 4 and the third equation by 3 to make the coefficients of c equal.

-1176 = 392a - 56b + 8c

-450 = 75a - 15b + 3c

Now, we can subtract the first equation from the second equation to eliminate c.

-726 = -317a + 41b

Now, we have two equations with two unknowns. We can solve this system using substitution or elimination. Let's use the substitution method.

First, we can solve the second equation for a.

a = (-726 + 317b) / 41

Now, we can substitute this expression for a into the first equation.

-60 = 10((-726 + 317b) / 41) - 4b

Now, we can solve for b.

-60 = (-7260 + 3170b) / 41 - 4b

-60(41) = -7260 + 3170b - 4b(41)

2460 = -7260 + 3170b - 164b

2460 + 7260 = 3170b - 164b

9720 = 3006b

b = 9720 / 3006

b = 3.22

Now, we can substitute this value of b into one of the original equations to solve for a.

a = (-726 + 317(3.22)) / 41

a = (-726 + 1020.74) / 41

a = 294.74 / 41

a = 7.19

Quadratic Equation

Now that we have the coefficients a, b, and c, we can write the quadratic equation as follows:

y = 7.19x^2 + 3.22x + 252

This equation represents the quadratic relationship between the variables x and y in the given table.

In this article, we explored the linear and quadratic relationships in a given table. We used the method of finding the slope and y-intercept to determine the linear relationship, and the method of finding the coefficients a, b, and c to determine the quadratic relationship. We found that the linear equation is y = 78x + 252, and the quadratic equation is y = 7.19x^2 + 3.22x + 252. These equations represent the relationships between the variables x and y in the given table.
Frequently Asked Questions (FAQs) about Linear and Quadratic Relationships

A: A linear relationship is a relationship between two variables where one variable is a constant multiple of the other variable. In other words, if we have two variables x and y, a linear relationship between them can be expressed as y = mx + b, where m is the slope of the line and b is the y-intercept. A quadratic relationship, on the other hand, is a relationship between two variables where one variable is a polynomial of degree two in the other variable. In other words, if we have two variables x and y, a quadratic relationship between them can be expressed as y = ax^2 + bx + c, where a, b, and c are constants.

A: To determine the linear relationship between two variables, you can use the method of finding the slope and y-intercept. The slope of a line is a measure of how much the line rises (or falls) vertically over a given horizontal distance. It can be calculated using the formula:

m = (y2 - y1) / (x2 - x1)

where (x1, y1) and (x2, y2) are two points on the line.

A: To determine the quadratic relationship between two variables, you can use the method of finding the coefficients a, b, and c. The coefficients can be found by substituting the values of x and y from the table into the equation y = ax^2 + bx + c and solving for a, b, and c.

A: The slope of a line is a measure of how much the line rises (or falls) vertically over a given horizontal distance. It can be used to determine the rate of change of the variable y with respect to the variable x.

A: The y-intercept of a line is the point where the line intersects the y-axis. It can be used to determine the value of the variable y when the variable x is equal to zero.

A: No, a quadratic relationship cannot be expressed as a linear relationship. A quadratic relationship is a polynomial of degree two, while a linear relationship is a polynomial of degree one.

A: Yes, a linear relationship can be expressed as a quadratic relationship. For example, the linear equation y = 2x + 3 can be expressed as the quadratic equation y = 2x^2 + 6x + 9.

A: To graph a linear or quadratic relationship, you can use a graphing calculator or a computer program. You can also use a coordinate plane to plot the points and draw the line or curve.

A: Linear and quadratic relationships have many real-world applications, including:

Physics: The motion of an object under the influence of gravity can be modeled using a quadratic relationship.
Economics: The demand for a product can be modeled using a linear relationship.
Engineering: The stress on a beam can be modeled using a quadratic relationship.

A: Yes, linear and quadratic relationships can be used to make predictions. For example, if you know the value of x and the corresponding value of y, you can use the equation to predict the value of y for a different value of x.

A: To determine the accuracy of a linear or quadratic relationship, you can use the method of least squares. This involves finding the best-fitting line or curve that minimizes the sum of the squared errors between the observed values and the predicted values.