Finding Quadratic Function With Lead Coefficient 3 And Roots 4 And 1

This article delves into the process of identifying a specific second-degree polynomial function, also known as a quadratic function, based on the given information. Specifically, we aim to find the quadratic function f(x) that satisfies two crucial conditions: it has a lead coefficient of 3 and possesses roots at x = 4 and x = 1. This exploration will involve understanding the relationship between roots, factors, and the general form of a quadratic function, ultimately leading us to the correct solution.

Understanding Quadratic Functions and Their Roots

At the heart of this problem lies the understanding of quadratic functions. A quadratic function is a polynomial function of degree two, generally expressed in the form f(x) = ax² + bx + c, where a, b, and c are constants, and a ≠ 0. The coefficient a is referred to as the lead coefficient, which plays a crucial role in determining the parabola's direction (upward if a > 0, downward if a < 0) and its vertical stretch or compression.

The roots of a quadratic function, also known as its zeros or x-intercepts, are the values of x for which f(x) = 0. These roots represent the points where the parabola intersects the x-axis. A fundamental property of quadratic functions is that they can have at most two real roots. These roots are intrinsically linked to the factors of the quadratic expression.

The Factor Theorem provides the bridge between roots and factors. It states that if r is a root of a polynomial function f(x), then (x - r) is a factor of f(x). Conversely, if (x - r) is a factor of f(x), then r is a root of f(x). This theorem is the cornerstone of constructing a quadratic function when its roots are known.

In our specific problem, we are given that the roots are 4 and 1. Applying the Factor Theorem, we can deduce that (x - 4) and (x - 1) are factors of the quadratic function f(x). This understanding forms the foundation for constructing the desired quadratic function.

Constructing the Quadratic Function

Given the roots 4 and 1, we know that (x - 4) and (x - 1) are factors of the quadratic function f(x). Therefore, we can express f(x) in the form:

f(x) = k(x - 4)(x - 1)

where k is a constant. This constant k represents a vertical stretch or compression factor and is directly related to the lead coefficient of the quadratic function. The lead coefficient is the coefficient of the x² term when the function is expanded.

We are given that the lead coefficient is 3. To determine the value of k, we need to expand the factored form of the quadratic function and equate the coefficient of the x² term to 3.

Expanding the expression, we get:

f(x) = k(x² - x - 4x + 4) f(x) = k(x² - 5x + 4)

f(x) = kx² - 5kx + 4k

Now, we can clearly see that the coefficient of the x² term is k. Since the lead coefficient is given as 3, we have:

k = 3

Substituting k = 3 back into the expression for f(x), we obtain:

f(x) = 3(x² - 5x + 4) f(x) = 3x² - 15x + 12

This is the quadratic function that satisfies the given conditions: a lead coefficient of 3 and roots at 4 and 1.

Verifying the Solution

To ensure the accuracy of our solution, it is crucial to verify that the function f(x) = 3x² - 15x + 12 indeed has roots at 4 and 1 and a lead coefficient of 3. The lead coefficient is readily apparent as the coefficient of the x² term, which is 3, confirming the first condition.

To verify the roots, we can substitute x = 4 and x = 1 into the function and check if the result is zero:

For x = 4:

f(4) = 3(4)² - 15(4) + 12 f(4) = 3(16) - 60 + 12 f(4) = 48 - 60 + 12 f(4) = 0

For x = 1:

f(1) = 3(1)² - 15(1) + 12 f(1) = 3 - 15 + 12 f(1) = 0

Both substitutions result in f(x) = 0, confirming that 4 and 1 are indeed the roots of the function. Therefore, our solution f(x) = 3x² - 15x + 12 satisfies all the given conditions.

Analyzing the Incorrect Options

Now, let's examine the incorrect options provided in the problem to understand why they do not satisfy the given conditions.

Option A: f(x) = 3x² + 5x + 4

This function has a lead coefficient of 3, which matches the given condition. However, if we try to factor this quadratic or use the quadratic formula to find its roots, we will find that it does not have roots at 4 and 1. Therefore, this option is incorrect.

Option B: f(x) = 3x² + 15x + 12

Again, this function has a lead coefficient of 3. However, similar to option A, its roots are not 4 and 1. We can factor out a 3 to get 3(x² + 5x + 4), which factors further into 3(x + 1)(x + 4). This indicates roots of -1 and -4, not 4 and 1. Thus, this option is incorrect.

Option C: f(x) = 3x² - 5x + 4

This function also has the correct lead coefficient of 3. However, its roots are not 4 and 1. Attempting to factor this quadratic or applying the quadratic formula will reveal that its roots are not the specified values. Therefore, this option is incorrect.

Conclusion: The Correct Quadratic Function

In conclusion, by applying the Factor Theorem and utilizing the given information about the lead coefficient and roots, we successfully constructed the second-degree polynomial function f(x) = 3x² - 15x + 12. This function satisfies the conditions of having a lead coefficient of 3 and roots at x = 4 and x = 1. The process involved understanding the relationship between roots and factors, expanding the factored form, and equating the lead coefficient to the given value. Verifying the solution by substituting the roots into the function confirmed its accuracy. Furthermore, analyzing the incorrect options highlighted the importance of satisfying both the lead coefficient and root conditions to arrive at the correct quadratic function.

This problem demonstrates a fundamental concept in algebra, emphasizing the connection between roots, factors, and the coefficients of a polynomial function. Understanding these relationships is crucial for solving a wide range of problems involving polynomials.