Polynomial Function: Find Equation With Zeros & Point

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Hey guys! Let's dive into a common yet fascinating problem in algebra: finding a polynomial function when we're given its zeros, degree, and a specific point it passes through. This might sound intimidating at first, but trust me, we'll break it down step by step so it's super clear. In this comprehensive guide, we'll explore how to construct a polynomial function using its zeros, degree, and a point it passes through. We'll walk through the process step by step, ensuring you grasp the underlying concepts and can confidently tackle similar problems. This is a crucial skill in algebra and calculus, so let's get started!

Understanding the Basics

Before we jump into the solution, let's quickly recap some fundamental concepts. Remember, a polynomial function is an expression consisting of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents. The zeros of a polynomial are the x-values where the function equals zero – in other words, the points where the graph crosses the x-axis. The degree of a polynomial is the highest power of the variable in the expression, and it tells us a lot about the shape and behavior of the graph. And lastly, a point on the graph, represented as (x, y), gives us a specific coordinate that the function must satisfy. Knowing these basics is super important for tackling our problem.

What is a Polynomial Function?

Let's kick things off by making sure we're all on the same page about what a polynomial function actually is. In simple terms, a polynomial function is a mathematical expression that combines variables (usually denoted as 'x'), constants (numbers), and exponents (positive whole numbers) through addition, subtraction, and multiplication. You've probably seen them before – things like x^2 + 3x - 4 or 5x^3 - 2x + 1 are classic examples. The highest exponent in the polynomial is what we call the degree of the polynomial, which plays a big role in determining the shape and behavior of its graph. For instance, a polynomial with a degree of 2 is a quadratic (think parabolas), while a degree of 3 gives us a cubic function (with its characteristic S-shape). Polynomials are super versatile and show up all over the place in math and science, so understanding them is key.

Zeros and Their Significance

Now, let's talk about zeros. The zeros of a polynomial function are the values of 'x' that make the function equal to zero. Graphically, these are the points where the polynomial's graph intersects or touches the x-axis. Zeros are also often called roots or solutions of the polynomial equation. Finding the zeros of a polynomial is a fundamental problem in algebra, and it has a ton of applications. For example, if you're modeling the trajectory of a projectile, the zeros might represent when the projectile hits the ground. Each zero corresponds to a factor of the polynomial. If 'r' is a zero, then (x - r) is a factor. This connection between zeros and factors is super important when we're trying to construct a polynomial function, as we'll see in our example. Knowing the zeros gives us the building blocks we need to write the polynomial in factored form.

Degree of the Polynomial

The degree of a polynomial is the highest power of the variable in the polynomial. For example, in the polynomial 3x^4 - 2x^2 + x - 5, the degree is 4 because the highest power of 'x' is x^4. The degree of a polynomial tells us a lot about its end behavior and the maximum number of turning points it can have. A polynomial of degree 'n' can have at most n-1 turning points (where the graph changes direction). The degree also affects the general shape of the graph. Linear functions (degree 1) are straight lines, quadratic functions (degree 2) are parabolas, and so on. In our problem, we're given that the polynomial has a degree of 3, which means we're dealing with a cubic function. This information helps us anticipate the general form of the polynomial and how many zeros it can have (a cubic can have up to 3 zeros).

The Role of a Point on the Graph

Finally, let's discuss the significance of a point on the graph. When we say a graph passes through a point (x, y), it means that if we plug in the x-value into the polynomial function, we should get the y-value as the output. In other words, the point (x, y) satisfies the equation of the polynomial. This is crucial because it gives us a specific condition that the polynomial must meet. In our problem, we're given the point (6, 144), which means that when x = 6, the polynomial function should equal 144. We'll use this information to find the leading coefficient of the polynomial, which is the final piece of the puzzle. Think of this point as a constraint that helps us nail down the exact polynomial function that fits all the given criteria.

Problem Breakdown: Zeros, Degree, and a Point

Okay, let's break down the specific problem we're tackling. We need to find a polynomial function that satisfies these conditions:

  • Zeros: -3, 2, and 4. These are the x-values where the polynomial equals zero.
  • Degree: 3. This tells us we're looking for a cubic polynomial (something of the form ax^3 + bx^2 + cx + d).
  • Point: (6, 144). This means the graph of the polynomial passes through the point where x = 6 and y = 144.

This might seem like a lot of information, but it's actually perfect for constructing our polynomial. The zeros will help us write the polynomial in factored form, the degree confirms the structure, and the point will allow us to determine the leading coefficient. Let's get started with the factored form!

Step 1: Using Zeros to Form Factors

The first step in finding our polynomial function is to use the zeros to create factors. Remember, if a number 'r' is a zero of a polynomial, then (x - r) is a factor of that polynomial. This is a fundamental concept in algebra and it's the key to unlocking our problem. We have three zeros: -3, 2, and 4. So, let's translate those into factors:

  • For the zero -3, the corresponding factor is (x - (-3)), which simplifies to (x + 3).
  • For the zero 2, the corresponding factor is (x - 2).
  • For the zero 4, the corresponding factor is (x - 4).

Now, we can write our polynomial in factored form. It will look something like this: f(x) = a(x + 3)(x - 2)(x - 4), where 'a' is a constant we need to determine. This constant is crucial because it scales the polynomial and ensures it passes through the given point. The factored form captures the zeros perfectly, but we still need to find 'a' to get the exact polynomial we're looking for. This is where the given point comes into play.

Step 2: Constructing the Polynomial in Factored Form

Now that we have our factors, let's construct the polynomial in its factored form. This is a super useful way to represent a polynomial when you know its zeros, because it directly incorporates that information. We know our polynomial, f(x), will look something like this:

f(x) = a(x + 3)(x - 2)(x - 4)

Where:

  • (x + 3), (x - 2), and (x - 4) are the factors we derived from the zeros.
  • a is the leading coefficient, which we still need to find. This coefficient will stretch or compress the polynomial vertically, and it's what makes our function unique and ensures it passes through the point (6, 144).

This factored form is a great starting point. It tells us a lot about the polynomial's behavior – namely, where it crosses the x-axis. But to nail down the exact function, we need to find the value of 'a'. This is where the point (6, 144) becomes our best friend. It gives us a specific input-output relationship that the polynomial must satisfy, which will allow us to solve for 'a'.

Step 3: Using the Given Point to Find the Leading Coefficient

This is where the point (6, 144) comes into play. Remember, this point tells us that when x = 6, f(x) = 144. We can use this information to solve for the leading coefficient 'a' in our factored form. Let's plug in x = 6 and f(x) = 144 into our equation:

144 = a(6 + 3)(6 - 2)(6 - 4)

Now, let's simplify this equation:

144 = a(9)(4)(2) 144 = a(72)

To solve for 'a', we'll divide both sides of the equation by 72:

a = 144 / 72 a = 2

So, we've found that the leading coefficient a = 2. This is a crucial step because it completes our polynomial function. We now know all the pieces: the factors from the zeros and the leading coefficient from the given point. Next, we'll put it all together to write the final polynomial function.

Step 4: Constructing the Final Polynomial Function

Alright, we've done the groundwork, and now it's time to put everything together and construct our final polynomial function. We've found the leading coefficient, a = 2, and we have the factored form of the polynomial:

f(x) = a(x + 3)(x - 2)(x - 4)

Now, let's plug in the value of 'a':

f(x) = 2(x + 3)(x - 2)(x - 4)

This is a perfectly valid form of the polynomial, and it clearly shows the zeros and the leading coefficient. However, sometimes it's useful to expand this into standard form (ax^3 + bx^2 + cx + d). Let's do that:

First, let's multiply the factors:

(x + 3)(x - 2) = x^2 + x - 6

Now, multiply the result by (x - 4):

(x^2 + x - 6)(x - 4) = x^3 - 3x^2 - 10x + 24

Finally, multiply the entire expression by the leading coefficient, 2:

2(x^3 - 3x^2 - 10x + 24) = 2x^3 - 6x^2 - 20x + 48

So, our final polynomial function in standard form is:

f(x) = 2x^3 - 6x^2 - 20x + 48

This is the polynomial function that has zeros at -3, 2, and 4, has a degree of 3, and passes through the point (6, 144). We did it!

Verification

Before we celebrate too much, let's just double-check that our polynomial actually satisfies all the given conditions. This is a great habit to get into, because it helps catch any small errors that might have crept in along the way.

  • Zeros: We can plug in x = -3, x = 2, and x = 4 into our polynomial and make sure the result is zero. Let's try x = -3:

    f(-3) = 2(-3)^3 - 6(-3)^2 - 20(-3) + 48 = -54 - 54 + 60 + 48 = 0 (✓)

    You can try the other zeros yourself, but they should also result in zero.

  • Point: Now let's check if the polynomial passes through (6, 144). We'll plug in x = 6:

    f(6) = 2(6)^3 - 6(6)^2 - 20(6) + 48 = 432 - 216 - 120 + 48 = 144 (✓)

Our polynomial checks out! It satisfies all the given conditions, which means we've successfully found the polynomial function.

Conclusion

Woohoo! We've successfully found the polynomial function that fits the given criteria. Finding a polynomial function given its zeros, degree, and a point can seem tricky at first, but by breaking it down into steps, it becomes much more manageable. We started by using the zeros to write the polynomial in factored form, then used the given point to solve for the leading coefficient. Finally, we constructed the polynomial in both factored and standard forms and verified our answer. Remember, the key is to understand the relationship between zeros and factors, and how a point on the graph can help you nail down the exact polynomial. Keep practicing, and you'll become a polynomial pro in no time!

I hope this guide has helped you understand the process clearly. Now you can confidently tackle similar problems. Keep practicing, and you'll master these types of questions in no time! Happy calculating, guys!