Hey guys! Ever wondered how to dissect a cubic function and uncover its hidden coefficients? Well, you've landed in the right spot! Today, we're diving deep into the fascinating world of cubic functions, specifically those in the form y = ax³ + bx² + cx + d. We're going to tackle a classic problem: How to determine the coefficients a, b, and c in terms of d when given points that the graph passes through. Buckle up, because we're about to embark on a mathematical adventure!
Understanding Cubic Functions and Their Graphs
Before we jump into the problem, let's solidify our understanding of cubic functions. A cubic function, at its heart, is a polynomial function of degree three. This means the highest power of the variable x is 3. The general form, as mentioned earlier, is y = ax³ + bx² + cx + d, where a, b, c, and d are constants, and importantly, a cannot be zero (otherwise, it wouldn't be cubic!). The graph of a cubic function is a curve that can have a variety of shapes, characterized by its twists and turns. It can have up to two turning points (where the graph changes direction) and can cross the x-axis up to three times, indicating the presence of real roots.
Key characteristics of cubic functions include:
- The coefficient a: This leading coefficient plays a crucial role in determining the end behavior of the graph. If a is positive, the graph rises to the right and falls to the left. Conversely, if a is negative, the graph falls to the right and rises to the left. This is something super useful to keep in mind when visualizing or sketching cubic functions.
- The constant term d: This term represents the y-intercept of the graph, the point where the graph intersects the y-axis. It's the value of y when x is zero. Knowing the y-intercept can give us a significant head start in understanding the graph's position.
- Turning Points: Cubic functions can have up to two turning points, which are local maxima or minima. These points are where the graph changes direction, going from increasing to decreasing or vice versa. The turning points significantly shape the curve of the graph.
- Roots or x-intercepts: A cubic function can have up to three real roots, which are the x-values where the graph crosses the x-axis (where y = 0). These roots are crucial for solving cubic equations and understanding the function's behavior.
Graphs of cubic functions can have a variety of shapes, but they typically exhibit an S-like curve or a variation thereof. They can have one, two, or three real roots, depending on how the curve intersects the x-axis. The shape and position of the graph are determined by the coefficients a, b, c, and d. Each coefficient influences the graph in its unique way, making the analysis of cubic functions quite intriguing.
Setting Up the Equations: Using Given Points
Now, let's get back to our main task. We're given that the graph of the cubic function y = ax³ + bx² + cx + d passes through three specific points: (-1, 6), (1, -2), and (3, 4). Our mission, should we choose to accept it (and we do!), is to find the coefficients a, b, and c in terms of d. This means we want to express a, b, and c as algebraic expressions involving d.
The key to cracking this problem lies in the fact that if a point lies on the graph of a function, its coordinates must satisfy the function's equation. In simpler terms, if we plug the x and y values of a point into the equation, the equation must hold true. This gives us a powerful tool to generate equations involving our unknown coefficients.
Let's apply this concept to our given points:
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Point (-1, 6): Substituting x = -1 and y = 6 into our cubic function equation, we get: 6 = a(-1)³ + b(-1)² + c(-1) + d Simplifying, we have: -a + b - c + d = 6
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Point (1, -2): Substituting x = 1 and y = -2, we get: -2 = a(1)³ + b(1)² + c(1) + d Simplifying, we have: a + b + c + d = -2
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Point (3, 4): Substituting x = 3 and y = 4, we get: 4 = a(3)³ + b(3)² + c(3) + d Simplifying, we have: 27a + 9b + 3c + d = 4
So, guys, we've successfully transformed our geometric problem into an algebraic one! We now have a system of three linear equations with four unknowns (a, b, c, and d):
- Equation 1: -a + b - c + d = 6
- Equation 2: a + b + c + d = -2
- Equation 3: 27a + 9b + 3c + d = 4
The fact that we have more unknowns than equations means we won't be able to find unique numerical values for a, b, and c. Instead, we'll express them in terms of d, which is exactly what the problem asks us to do.
Solving the System of Equations: Expressing a, b, and c in Terms of d
Now comes the fun part: solving our system of equations! We have several methods at our disposal, such as substitution, elimination, or matrix methods. For this problem, the elimination method seems particularly efficient. Our goal is to strategically eliminate variables until we isolate a, b, and c in terms of d.
Step 1: Eliminate c
Let's start by eliminating c from our equations. Notice that Equations 1 and 2 have -c and +c terms, making them perfect candidates for elimination. If we add these two equations together, the c terms will cancel out:
(-a + b - c + d) + (a + b + c + d) = 6 + (-2)
Simplifying, we get:
2b + 2d = 4
Dividing both sides by 2, we obtain a simpler equation:
b + d = 2
This gives us a direct expression for b in terms of d:
b = 2 - d
Awesome! We've already found one coefficient in terms of d. Now, let's keep going to find a and c.
Step 2: Another Equation Without c
To eliminate c again, let's manipulate Equations 2 and 3. To make the c terms have opposite coefficients, we can multiply Equation 2 by -3:
-3(a + b + c + d) = -3(-2)
This gives us:
-3a - 3b - 3c - 3d = 6
Now, we can add this modified equation to Equation 3:
(27a + 9b + 3c + d) + (-3a - 3b - 3c - 3d) = 4 + 6
Simplifying, we get:
24a + 6b - 2d = 10
Dividing both sides by 2, we have:
12a + 3b - d = 5
Step 3: Solve for a
We now have an equation with a, b, and d. But remember, we already have an expression for b in terms of d (b = 2 - d). Let's substitute this into our equation:
12a + 3(2 - d) - d = 5
Expanding and simplifying, we get:
12a + 6 - 3d - d = 5
12a - 4d = -1
Now, we can solve for a:
12a = 4d - 1
a = (4d - 1) / 12
Fantastic! We've found a in terms of d as well.
Step 4: Solve for c
We're in the home stretch now! To find c, we can use any of our original equations. Equation 2 (a + b + c + d = -2) looks like a good choice. We already have expressions for a and b in terms of d, so let's substitute them in:
((4d - 1) / 12) + (2 - d) + c + d = -2
To simplify, let's multiply the entire equation by 12 to get rid of the fraction:
(4d - 1) + 12(2 - d) + 12c + 12d = -24
Expanding and simplifying, we get:
4d - 1 + 24 - 12d + 12c + 12d = -24
12c + 4d + 23 = -24
12c = -4d - 47
c = (-4d - 47) / 12
And there we have it! We've successfully expressed c in terms of d.
Putting It All Together: The Coefficients in Terms of d
Okay, let's recap our findings. We started with a cubic function y = ax³ + bx² + cx + d and three points that the graph passes through. Through careful algebraic manipulation and the elimination method, we've determined the coefficients a, b, and c in terms of d:
- a = (4d - 1) / 12
- b = 2 - d
- c = (-4d - 47) / 12
These are our final answers! We've successfully solved the problem. Depending on the value of d, we can now find the corresponding values of a, b, and c.
Conclusion: The Power of Algebraic Manipulation
This problem beautifully illustrates the power of algebraic manipulation in solving mathematical problems. By translating the geometric information (points on a graph) into algebraic equations and then skillfully manipulating those equations, we were able to unravel the relationships between the coefficients of the cubic function. It's like detective work, but with numbers and symbols! Understanding cubic functions and their properties is crucial in various fields, including engineering, physics, and computer graphics. The ability to determine coefficients based on given points is a valuable skill.
So, guys, next time you encounter a cubic function problem, remember the techniques we've discussed here. Break down the problem, set up the equations, and use your algebraic prowess to find the solution. You've got this!