Find The Equation Of The Quadratic Function Whose Graph Is A Parabola Containing The Points $(-3,43), (1,3),$ And $(2,13$\].The Equation Of The Quadratic Function Is $y = \square$. (Type An Expression Using $x$ As The

Introduction

In mathematics, a quadratic function is a polynomial function of degree two, which means the highest power of the variable is two. The general form of a quadratic function is $y = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants. In this article, we will discuss how to find the equation of a quadratic function whose graph is a parabola containing three given points.

The Problem

We are given three points on the graph of a parabola: $(-3,43), (1,3),$ and $(2,13)$. Our goal is to find the equation of the quadratic function that passes through these three points.

The Method

To find the equation of the quadratic function, we will use the fact that the graph of a quadratic function is a parabola. A parabola is a U-shaped curve that can be described by a quadratic equation. We will use the three given points to set up a system of equations, which we will then solve to find the values of $a$, $b$, and $c$.

Step 1: Write the Equation of the Quadratic Function

The general form of a quadratic function is $y = ax^2 + bx + c$. We will use this equation to write three equations, one for each of the given points.

• For the point $(-3,43)$, we have: $43 = a(-3)^2 + b(-3) + c$
• For the point $(1,3)$, we have: $3 = a(1)^2 + b(1) + c$
• For the point $(2,13)$, we have: $13 = a(2)^2 + b(2) + c$

Step 2: Simplify the Equations

We will simplify each of the three equations by expanding the squared term and combining like terms.

• For the point $(-3,43)$, we have: $43 = 9a - 3b + c$
• For the point $(1,3)$, we have: $3 = a + b + c$
• For the point $(2,13)$, we have: $13 = 4a + 2b + c$

Step 3: Solve the System of Equations

We will solve the system of three equations to find the values of $a$, $b$, and $c$. We can use any method we like, such as substitution or elimination.

Let's use the elimination method to solve the system of equations. We will multiply the first equation by 2 and the second equation by 3 to make the coefficients of $b$ equal.

• For the point $(-3,43)$, we have: $86 = 18a - 6b + 2c$
• For the point $(1,3)$, we have: $9 = 3a + 3b + 3c$

Now, we will subtract the second equation from the first equation to eliminate the variable $b$.

$86 - 9 = (18a - 6b + 2c) - (3a + 3b + 3c)$

Simplifying the equation, we get:

$77 = 15a - 9b - c$

Now, we will multiply the third equation by 3 and the second equation by 4 to make the coefficients of $b$ equal.

• For the point $(2,13)$, we have: $39 = 12a + 6b + 3c$
• For the point $(1,3)$, we have: $12 = 4a + 4b + 4c$

Now, we will subtract the second equation from the first equation to eliminate the variable $b$.

$39 - 12 = (12a + 6b + 3c) - (4a + 4b + 4c)$

Simplifying the equation, we get:

$27 = 8a + 2b - c$

Now, we have two equations with two variables. We can solve this system of equations using substitution or elimination.

Let's use the substitution method to solve the system of equations. We will solve the first equation for $b$ and substitute it into the second equation.

$77 = 15a - 9b - c$

Solving for $b$, we get:

$b = \frac{15a - c - 77}{9}$

Now, we will substitute this expression for $b$ into the second equation.

$27 = 8a + 2(\frac{15a - c - 77}{9}) - c$

Simplifying the equation, we get:

$27 = 8a + \frac{30a - 2c - 154}{9} - c$

Multiplying both sides of the equation by 9, we get:

$243 = 72a + 30a - 2c - 154 - 9c$

Simplifying the equation, we get:

$243 = 102a - 11c - 154$

Now, we will add 154 to both sides of the equation to get:

$397 = 102a - 11c$

Now, we will add 11c to both sides of the equation to get:

$397 + 11c = 102a$

Now, we will divide both sides of the equation by 102 to get:

$\frac{397 + 11c}{102} = a$

Now, we will substitute this expression for $a$ into one of the original equations to solve for $c$.

Let's use the first equation.

$43 = a(-3)^2 + b(-3) + c$

Substituting the expression for $a$, we get:

$43 = (\frac{397 + 11c}{102})(-3)^2 + b(-3) + c$

Simplifying the equation, we get:

$43 = \frac{397 + 11c}{102} \cdot 9 - 3b + c$

Multiplying both sides of the equation by 102, we get:

$4374 = 397 \cdot 9 + 99c - 306b + 102c$

Simplifying the equation, we get:

$4374 = 3573 + 201c - 306b$

Now, we will add 306b to both sides of the equation to get:

$4374 + 306b = 3573 + 201c$

Now, we will subtract 3573 from both sides of the equation to get:

$801 = 306b + 201c$

Now, we will add 201c to both sides of the equation to get:

$801 + 201c = 306b$

Now, we will divide both sides of the equation by 306 to get:

$\frac{801 + 201c}{306} = b$

Now, we have expressions for $a$ and $b$ in terms of $c$. We can substitute these expressions into one of the original equations to solve for $c$.

Let's use the first equation.

$43 = a(-3)^2 + b(-3) + c$

Substituting the expressions for $a$ and $b$, we get:

$43 = (\frac{397 + 11c}{102})(-3)^2 + (\frac{801 + 201c}{306})(-3) + c$

Simplifying the equation, we get:

$43 = \frac{397 + 11c}{102} \cdot 9 - \frac{801 + 201c}{306} \cdot 3 + c$

Multiplying both sides of the equation by 306, we get:

$13068 = (397 + 11c) \cdot 9 \cdot 3 - (801 + 201c) \cdot 3 + 306c$

Simplifying the equation, we get:

$13068 = 10773 + 297c - 2403 - 603c + 306c$

Now, we will add 603c to both sides of the equation to get:

$13068 = 10773 + 297c + 603c - 2403 + 306c$

Now, we will subtract 10773 from both sides of the equation to get:

$2295 = 1206c - 1335$

Now, we will add 1335 to both sides of the equation to get:

$3629 = 1206c$

Now, we will divide both sides of the equation by 1206 to get:

$\frac{3629}{1206} = c$

Now, we have found the value of $c$. We can substitute this value into the expressions for $a$ and $b$ to find their values.

$a = \frac{397 + 11c}{102}$

$b = \frac{801 + 201c}{306}$

Substituting the value of $c$, we get:

$a = \frac{397 + 11 \cdot \frac{3629}{1206}}{102}$

$b = \frac{801 + 201 \cdot \frac{3629}{1206}}{306}$

Simplifying the expressions, we get:

$a = \frac{397 + \frac{39919}{1206}}{102}$

$b = \frac{801 + \frac{727969}{1206}}{306}$

Now, we will simplify the expressions further.

Q: What is a quadratic function?

A: A quadratic function is a polynomial function of degree two, which means the highest power of the variable is two. The general form of a quadratic function is $y = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants.

Q: How do I find the equation of a quadratic function given three points?

A: To find the equation of a quadratic function given three points, you can use the method of substitution or elimination. First, write the equation of the quadratic function in the form $y = ax^2 + bx + c$. Then, substitute the coordinates of each point into the equation to get a system of three equations. Solve the system of equations to find the values of $a$, $b$, and $c$.

Q: What is the significance of the coefficient $a$ in a quadratic function?

A: The coefficient $a$ in a quadratic function determines the direction and width of the parabola. If $a$ is positive, the parabola opens upward. If $a$ is negative, the parabola opens downward. The value of $a$ also affects the width of the parabola.

Q: How do I determine the vertex of a quadratic function?

A: To determine the vertex of a quadratic function, you can use the formula $x = -\frac{b}{2a}$. This formula gives the x-coordinate of the vertex. To find the y-coordinate of the vertex, substitute the x-coordinate into the equation of the quadratic function.

Q: What is the significance of the vertex of a quadratic function?

A: The vertex of a quadratic function is the highest or lowest point on the graph of the function. It is also the point of symmetry of the graph.

Q: How do I determine the x-intercepts of a quadratic function?

A: To determine the x-intercepts of a quadratic function, you can set the equation of the function equal to zero and solve for $x$. The x-intercepts are the points where the graph of the function crosses the x-axis.

Q: What is the significance of the x-intercepts of a quadratic function?

A: The x-intercepts of a quadratic function are the points where the graph of the function crosses the x-axis. They are also the solutions to the equation of the function.

Q: How do I determine the y-intercept of a quadratic function?

A: To determine the y-intercept of a quadratic function, you can substitute $x = 0$ into the equation of the function. The y-intercept is the point where the graph of the function crosses the y-axis.

Q: What is the significance of the y-intercept of a quadratic function?

A: The y-intercept of a quadratic function is the point where the graph of the function crosses the y-axis. It is also the value of the function when $x = 0$.

Q: Can I use a graphing calculator to find the equation of a quadratic function?

A: Yes, you can use a graphing calculator to find the equation of a quadratic function. First, enter the coordinates of the three points into the calculator. Then, use the calculator's built-in functions to find the equation of the quadratic function.

Q: What are some common applications of quadratic functions?

A: Quadratic functions have many common applications in mathematics, science, and engineering. Some examples include:

• Modeling the motion of objects under the influence of gravity
• Finding the maximum or minimum value of a function
• Determining the area or perimeter of a shape
• Solving problems involving optimization

Q: Can I use quadratic functions to model real-world problems?

A: Yes, you can use quadratic functions to model real-world problems. Quadratic functions can be used to model a wide range of phenomena, including the motion of objects, the growth of populations, and the behavior of electrical circuits.

Q: What are some common mistakes to avoid when working with quadratic functions?

A: Some common mistakes to avoid when working with quadratic functions include:

• Failing to check the domain of the function
• Failing to check the range of the function
• Failing to consider the vertex of the function
• Failing to consider the x-intercepts of the function

By following these tips and avoiding common mistakes, you can master the art of working with quadratic functions and apply them to a wide range of real-world problems.