Problem 53 - Algebra and Trigonometry

Question

Polynomials Determined by a Set of Points We all know that two points uniquely determine a line \(y=a x+b\) in the coordinate plane. Similarly, three points uniquely determine a quadratic (second-degree) polynomial $$y=a x^{2}+b x+c$$ four points uniquely determine a cubic (third-degree) polynomial $$y=a x^{3}+b x^{2}+c x+d$$ and so on. (Some exceptions to this rule are if the three points actually lic on a line, or the four points lie on a quadratic or line, and so on.) For the following set of five points, find the line that contains the first two points, the quadratic that contains the first three points, the cubic that contains the first four points, and the fourth-degree polynomial that contains all five points. $$(0,0),(1,12),(2,40),(3,6),(-1,-14)$$ Graph the points and functions in the same viewing rectangle using a graphing device.

Step-by-Step Solution

Answer

The line is \(y = 12x\), the quadratic is \(y = 6x^2 + 6x\), the cubic is \(y = -x^3 + 8x^2 + 6x\), and the quartic is \(y = x^4 - 6x^3 + 11x^2 + 6x\).

Step 1: Identify the Line from the First Two Points

To find the line through the first two points \((0,0)\) and \((1,12)\), determine the slope \(m\) and then the line equation. The slope is found as \(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{12 - 0}{1 - 0} = 12\). The line can be expressed as \(y = mx + b\). Substituting \(b = 0\) since the line passes through the origin, we get the line equation \(y = 12x\).

Step 2: Find the Quadratic through the First Three Points

The quadratic polynomial is in the form \(y = ax^2 + bx + c\). We use the points \((0,0)\), \((1,12)\), and \((2,40)\) to create a system of equations:\[\begin{align*}0 &= a(0)^2 + b(0) + c, \12 &= a(1)^2 + b(1) + c, \40 &= a(2)^2 + b(2) + c.\end{align*}\].\From the first equation, \(c = 0\). Substituting in the other equations gives the system:\[\begin{align*}12 &= a + b, \40 &= 4a + 2b.\end{align*}\].\Solve these to find \(a = 6\) and \(b = 6\). So, the quadratic is \(y = 6x^2 + 6x\).

Step 3: Find the Cubic through the First Four Points

For the cubic \(y = ax^3 + bx^2 + cx + d\), utilize the points \((0,0)\), \((1,12)\), \((2,40)\), and \((3,6)\) to generate equations. The system is: \[\begin{align*}0 &= 0, \12 &= a + b + c + d, \40 &= 8a + 4b + 2c + d, \6 &= 27a + 9b + 3c + d\end{align*}\].\ Solving, we find \(a = -1, b = 8, c = 6, d = 0\). Thus, the cubic polynomial is \(y = -x^3 + 8x^2 + 6x\).

Step 4: Determine the Fourth-Degree Polynomial through All Five Points

For the polynomial \(y = ax^4 + bx^3 + cx^2 + dx + e\) through \((0,0)\), \((1,12)\), \((2,40)\), \((3,6)\), and \((-1,-14)\), set up the equations:\[\begin{align*}0 &= e, \12 &= a + b + c + d + e, \40 &= 16a + 8b + 4c + 2d + e, \6 &= 81a + 27b + 9c + 3d + e, \-14 &= a - b + c - d + e\end{align*}\].\Solving this system results in \(a = 1, b = -6, c = 11, d = 6, e = 0\). Thus, the polynomial is \(y = x^4 - 6x^3 + 11x^2 + 6x\).

Step 5: Graph the Points and Functions

Use a graphing device to plot the points \((0,0), (1,12), (2,40), (3,6), (-1,-14)\) and overlay the functions \(y = 12x\), \(y = 6x^2 + 6x\), \(y = -x^3 + 8x^2 + 6x\), and \(y = x^4 - 6x^3 + 11x^2 + 6x\) on the same coordinate system. Adjust the viewing window so that all points and curves are visible.

Key Concepts

Quadratic Equations

Quadratic equations are the foundation of many algebraic concepts and are important in describing the motion of objects and shapes of graphs. A quadratic polynomial is generally expressed as: \[ y = ax^2 + bx + c \]. The graph of a quadratic equation is a parabola, which can open upwards or downwards depending on the coefficient \(a\).

• If \(a > 0\), the parabola opens upwards, resembling a 'U' shape.
• If \(a < 0\), the parabola opens downwards, resembling an inverted 'U'.
• The vertex of the parabola is the point where it changes direction, and this can be found using the vertex formula \( x = \frac{-b}{2a} \).

For our specific problem, we used the points (0,0), (1,12), and (2,40) to create a quadratic equation. By formulating equations based on these points and solving them, we determined that the quadratic polynomial is \( y = 6x^2 + 6x \). This parabola opens upwards since the coefficient of \( x^2 \) is positive (\(a = 6\)), and its vertex can be calculated using the mentioned formula. This shows the practical application of quadratic equations in forming polynomials to fit a set of points.

Cubic Polynomials

Cubic polynomials extend the concepts found in linear and quadratic equations, introducing a third-degree term \(x^3\) that adds more complexity and flexibility to the graph. The general form of a cubic polynomial is: \[ y = ax^3 + bx^2 + cx + d \]. Cubic equations often have an "S" shaped curve, which can intersect the x-axis up to three times, giving the potential for three real roots.

• These polynomials can model changes and movements more dynamically because they can describe bends and inflections in the graph.
• The number and nature of the roots (or solutions) depend greatly on the coefficients.

In our problem, the cubic polynomial incorporates the points (0,0), (1,12), (2,40), and (3,6), leading to the equation \( y = -x^3 + 8x^2 + 6x \). By solving the system of equations formed by these points, we identified the coefficients \(a = -1, b = 8, c = 6, d = 0\), which tell us the polynomial's behavior and shape. This polynomial showcases the ability of cubic functions to create complex curvatures and intersections, capturing the unique characteristics between chosen data points.

Graphing Functions

Graphing functions is a powerful method to visually analyze and interpret polynomial behaviors and their relationships with data points. It involves plotting equations on a coordinate plane to observe intersections, curves, and shifts.

• A graph provides a snapshot of a function's behavior at different inputs.
• It helps identify key features like roots, intersections, maxima, minima, and asymptotic behavior.

In this exercise, after determining each polynomial expression for different sets of points, graphing provides an immediate visual interpretation. We plotted the points (0,0), (1,12), (2,40), (3,6), (-1,-14) and the polynomials:

• Line: \(y = 12x\)
• Quadratic: \(y = 6x^2 + 6x\)
• Cubic: \(y = -x^3 + 8x^2 + 6x\)
• Fourth-degree: \(y = x^4 - 6x^3 + 11x^2 + 6x\)

Using a graphing tool in the same viewing rectangle ensures all functions can be reviewed together, allowing for a comparison of how each polynomial fits its respective points and how they interact across the curve. This practice highlights how graphing is an essential tool for understanding polynomial solutions and predicting their behavior in real-world applications.