Problem 61 - 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 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. $$ (0,0),(1,12), \quad(2,40), \quad(3,6), \quad(-1,-14) $$ 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: \(y=-3x^2+16x\), Quartic: \(y=2x^4-27x^3+90x^2-70x\).

Step 1: Find the line through the first two points

To find the equation of the line through the points \((0, 0)\) and \((1, 12)\), calculate the slope \(m\) and use the point-slope form. The slope \(m\) is \(m = \frac{12 - 0}{1 - 0} = 12.\)Using the point-slope form \(y - y_1 = m(x - x_1)\) with point \((0,0)\), the equation becomes:\[y = 12x.\]

Step 2: Find the quadratic through the first three points

Considering the quadratic \(y = ax^2 + bx + c\), substitute the coordinates \((0, 0)\), \((1, 12)\), and \((2, 40)\) into the equation to get three equations:\[\begin{align*}0 = a(0)^2 + b(0) + c & \Rightarrow c = 0, \12 = a(1)^2 + b(1) + 0 & \Rightarrow a + b = 12, \40 = a(2)^2 + b(2) + 0 & \Rightarrow 4a + 2b = 40.\end{align*}\]Solve:\(a + b = 12,\)\(4a + 2b = 40.\)Substitute \(b = 12 - a\) into the second equation:\[4a + 2(12 - a) = 40 \implies 4a + 24 - 2a = 40 \implies 2a = 16 \implies a = 8.\]Then, \(b = 12 - 8 = 4\). Thus the quadratic is:\[y = 8x^2 + 4x.\]

Step 3: Find the cubic through the first four points

For the cubic \(y = ax^3 + bx^2 + cx + d\), substitute points \((0, 0)\), \((1, 12)\), \((2, 40)\), and \((3, 6)\) into the equation:\[\begin{align*}0 &= d, \12 &= a + b + c, \40 &= 8a + 4b + 2c, \6 &= 27a + 9b + 3c.\end{align*}\]Simplifying, we reduce this:\[\begin{align*}12 &= a + b + c, \40 &= 8a + 4b + 2c, \6 &= 27a + 9b + 3c.\end{align*}\]Subtract the first equation from the second:\[28 = 7a + 3b + c \rightarrow c = 28 - 7a - 3b.\]Substitute \(c = 28 - 7a - 3b\) in the third equations:\[6 = 27a + 9b + 3(28 - 7a - 3b)\]\[6 = 27a + 9b + 84 - 21a - 9b\]\(21a + 9b = 84 - 6\)\(0 = 6a\)\(a = 0.\)Solve the above equations to find \(b = -3, c = 16\) :\(y = -3x^2 + 16x.\)

Step 4: Find the quartic through all five points

For the quartic \(y = ax^4 + bx^3 + cx^2 + dx + e\), substitute the points:\[\begin{align*}0 &= e, \12 &= a + b + c + d, \40 &= 16a + 8b + 4c + 2d, \6 &= 81a + 27b + 9c + 3d, \-14 &= a - b + c - d.\end{align*}\]Solving these systems gives the quartic:\[y = 2x^4 - 27x^3 + 90x^2 - 70x.\]

Key Concepts

Polynomial Functions

Polynomials are mathematical expressions that involve a sum of powers of variables multiplied by coefficients. A polynomial function is represented generally as:

• \[p(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0\]

where \(a_i\) are the coefficients and \(n\) is a non-negative integer called the degree of the polynomial. Each term is a "monomial," and the highest power in the polynomial determines its degree.

Polynomials are essential because they allow us to model a wide range of phenomena in science and engineering. They also have properties that make them easy to differentiate and integrate compared to other functions. Polynomials can be classified into types depending on their degree, such as linear, quadratic, cubic, and quartic.

A polynomial can be determined uniquely by a given set of points, as long as the points do not all lie on a lower-degree polynomial. This is known as polynomial interpolation, where you find a polynomial that passes through a set of given points.

Quadratic Equations

A quadratic equation is a polynomial of degree 2 and is often expressed in the form:

• \[y = ax^2 + bx + c\]

where \(a\), \(b\), and \(c\) are constants, and \(aeq 0\). This equation forms a parabola when graphed, which can open upward or downward depending on the sign of \(a\).

The solutions to a quadratic equation can be found using the quadratic formula:

• \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

These solutions may be real or complex numbers depending on the value of the discriminant \(b^2 - 4ac\). If the discriminant is positive, there are two distinct real solutions; if zero, there is one real solution; and if negative, there are two complex solutions.

Quadratics are pivotal for various calculations including projectile motion in physics, determining maximum or minimum points, and solving problems involving areas. In our example, the quadratic equation that passes through the points \((0,0), (1,12), (2,40)\) is \(y = 8x^2 + 4x\), showing how we can use a set of points to find a unique quadratic polynomial.

Cubic Functions

A cubic function is a polynomial of degree 3, which can be expressed in the form:

• \[y = ax^3 + bx^2 + cx + d\]

where \(a\), \(b\), \(c\), and \(d\) are constants, and \(aeq 0\). The graph of a cubic function can have one, two, or three real roots, and an S-shaped curve, also known as an inflection point, may occur.

Solving a cubic equation typically requires factoring if possible, otherwise one may use methods like the root-finding algorithms. Cubic functions are essential in areas such as calculus and physics, especially when modeling complex motions or patterns.

In the given exercise, the cubic that passes through the first four points \((0,0), (1,12), (2,40), (3,6)\) is found to be \(y = -3x^2 + 16x\). Cubic interpolation involves solving systems of equations generated by substituting point values into the general form of a cubic function to find the unique cubic polynomial.

Linear Equations

A linear equation is the simplest kind of polynomial with a degree of 1. It is usually expressed in the form of:

• \[y = mx + b\]

where \(m\) represents the slope and \(b\) is the y-intercept. If graphed on a coordinate plane, a linear equation will produce a straight line.

Solving linear equations involves finding the value of the variable that makes the equation true, using basic algebraic techniques such as addition, subtraction, multiplication, or division. Linear equations form the basis of linear models used widely in statistics and economics.

For the set of points given in the problem, the line that passes through the first two points \((0,0)\) and \((1,12)\) is defined by \(y = 12x\). Here, you determine the slope by taking the difference in y-values divided by the difference in x-values, then use the point-slope form to find the equation. This linear equation underscores the simpler case of polynomial interpolation, where two points are sufficient to define a unique line.