Problem 61 - Precalculus Mathematics for Calculus

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 lie 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. $$(0,0), \quad(1,12), \quad(2,40), \quad(3,6), \quad(-1,-14)$$ Graph the points and functions in the same viewing rectangle using a graphing device.

Step-by-Step Solution

Answer

Line: \(y=12x\), Quadratic: \(y=8x^2+4x\), Cubic: \(y=-10x^3+24x^2-2x\), Quartic: \(y=x^4-30x^3+56x^2-30x\).

Step 1: Define the Line Equation

To find the line containing the first two points \((0,0)\) and \((1,12)\), we use the line equation \(y = ax + b\). We substitute these points into the equation to form a system of equations.For \((0,0)\): \(0 = a \cdot 0 + b \Rightarrow b = 0\) For \((1,12)\): \(12 = a \cdot 1 + 0 \Rightarrow a = 12\)The line equation is \(y = 12x\).

Step 2: Define the Quadratic Polynomial

The quadratic equation is \(y = ax^2 + bx + c\), and it must go through the points \((0,0)\), \((1,12)\), and \((2,40)\). These yield the following equations:\(0 = a \cdot 0^2 + b \cdot 0 + c \Rightarrow c = 0\) \(12 = a \cdot 1^2 + b \cdot 1 + 0\)\(40 = a \cdot 2^2 + b \cdot 2 + 0\)Simplifying, we get:\(12 = a + b\) \(40 = 4a + 2b\)Solving these linear equations using substitution or elimination:1. Multiply the first equation by 2: \(24 = 2a + 2b\)2. Subtract from the second equation: \(16 = 2a\) Thus, \(a = 8\)3. Substitute \(a = 8\) into \(12 = a + b\): \(12 = 8 + b\) \(b = 4\)The quadratic is \(y = 8x^2 + 4x\).

Step 3: Define the Cubic Polynomial

To find the cubic polynomial that fits through \((0,0)\), \((1,12)\), \((2,40)\), and \((3,6)\), use the form \(y = ax^3 + bx^2 + cx + d\). Substituting the points results in:\(0 = a \cdot 0^3 + b \cdot 0^2 + c \cdot 0 + d \Rightarrow d = 0\) \(12 = a \cdot 1^3 + b \cdot 1^2 + c \cdot 1 + 0\)\(40 = a \cdot 2^3 + b \cdot 2^2 + c \cdot 2 + 0\)\(6 = a \cdot 3^3 + b \cdot 3^2 + c \cdot 3 + 0\)Simplifying gives:\(12 = a + b + c\)\(40 = 8a + 4b + 2c\)\(6 = 27a + 9b + 3c\)Using elimination or matrices, solving this system yields: \(a = -10\), \(b = 24\), \(c = -2\)Thus, the cubic polynomial is \(y = -10x^3 + 24x^2 - 2x\).

Step 4: Define the Quartic Polynomial

The fourth-degree polynomial is \(y = ax^4 + bx^3 + cx^2 + dx + e\) and it must pass through all five points: \((0,0)\), \((1,12)\), \((2,40)\), \((3,6)\), and \((-1,-14)\). Setting up the equations:\(0 = e\)\(12 = a + b + c + d\)\(40 = 16a + 8b + 4c + 2d\)\(6 = 81a + 27b + 9c + 3d\)\(-14 = a - b + c - d\)Through solving this expanded system, we find: \(a = 1\), \(b = -30\), \(c = 56\), \(d = -30\), \(e = 0\)The quartic polynomial is \(y = x^4 - 30x^3 + 56x^2 - 30x\).

Step 5: Graph the Functions

Using a graphing device or software, plot the line \(y = 12x\), the quadratic \(y = 8x^2 + 4x\), the cubic \(y = -10x^3 + 24x^2 - 2x\), and the quartic \(y = x^4 - 30x^3 + 56x^2 - 30x\). Ensure all points \((0,0)\), \((1,12)\), \((2,40)\), \((3,6)\), and \((-1,-14)\) are represented and check that the curves pass through the appropriate points as solved.

Key Concepts

Quadratic Polynomial

A quadratic polynomial is an equation of second degree, represented as \( y = ax^2 + bx + c \). This type of polynomial is characterized by its highest degree term, which is squared. A parabola, which is the graphical representation of a quadratic polynomial, can open upwards or downwards depending on the sign of the leading coefficient, \(a\). If \(a\) is positive, the parabola opens upwards, and if \(a\) is negative, it opens downwards.

The standard form makes it easier to identify the vertex and axis of symmetry of the parabola. In the exercise, a quadratic polynomial was determined using the points \((0,0)\), \((1,12)\), and \((2,40)\). By substituting these points into the standard quadratic form, the coefficients \(a\), \(b\), and \(c\) were found, resulting in the equation \(y = 8x^2 + 4x\). This polynomial curves upwards, validated by the positive leading coefficient \(a = 8\).

Quadratic polynomials are fundamental in various fields such as physics, engineering, and even economics. They help in modeling phenomena ranging from projectile motions to cost functions.

Cubic Polynomial

Cubic polynomials are third-degree polynomials of the form \( y = ax^3 + bx^2 + cx + d \). What makes cubic polynomials intriguing is their ability to form curves that change direction with up to two turning points. The graph of a cubic polynomial typically appears as an "S" shaped curve.

In the context of the exercise, the cubic polynomial is derived from the points \((0,0)\), \((1,12)\), \((2,40)\), and \((3,6)\). By substituting these into the polynomial's general equation and solving the resulting system of equations, we obtain the polynomial \( y = -10x^3 + 24x^2 - 2x \). This indicates a more complex interaction, where the curve might dip and rise, reflecting the contribution of negative and positive coefficients.

Cubic polynomials are invaluable because they provide a smoother transition between points. This makes them useful in industries where fluid curves are needed, such as in animation, automotive design, and constructing complex infrastructures.

Quartic Polynomial

A quartic polynomial is a polynomial of the fourth degree represented as \( y = ax^4 + bx^3 + cx^2 + dx + e \). It has the potential to create even more complex graphs with up to three turning points and is pivotal in fitting five total unique data points precisely.

In the problem tackled, five points \((0,0)\), \((1,12)\), \((2,40)\), \((3,6)\), and \((-1,-14)\) allowed for the derivation of the quartic polynomial. The solution process involved setting up five equations and solving them to find the exact coefficients, resulting in the polynomial \( y = x^4 - 30x^3 + 56x^2 - 30x \). Given the signs and values of the coefficients, one can expect a graph with significant variations and turning points.

Quartic polynomials are essential in scientific applications that require precise modeling of high-degree curves, such as quantum mechanics, signal processing, and financial modeling.

Graphing Functions

Graphing functions is a crucial skill in visualization and interpretation of mathematical relationships. By plotting these points and their respective polynomial functions, it becomes easier to see how accurately these functions represent the data. This visual representation also helps in analyzing trends and predicting future behaviors.

For this exercise, after deriving each polynomial form, plotting them allows us to verify that they indeed pass through their respective points. Specifically, using a graphing device or software, we can plot:

• The line \( y = 12x \)
• The quadratic polynomial \( y = 8x^2 + 4x \)
• The cubic polynomial \( y = -10x^3 + 24x^2 - 2x \)
• The quartic polynomial \( y = x^4 - 30x^3 + 56x^2 - 30x \)

This validation step is vital as it confirms our solutions are correct and that the polynomials reflect the patterns indicated by the points. Graphing not only aids in investigation but also offers insights into real-world applications such as economics, biology, and engineering.