Question

Find a polynomial \(p(x)\) of degree 2 such that: $$ \text { a. } p(0)=2, p(1)=3, p(3)=8 $$ b. \(p(0)=5, p(1)=3, p(2)=5\)

Short answer

(a) \(p(x) = \frac{1}{2}x^2 + \frac{1}{2}x + 2\); (b) \(p(x) = 2x^2 - 4x + 5\).

Step-by-step solution

Identifying the General Form

A polynomial of degree 2 can be represented as \( p(x) = ax^2 + bx + c \). Our task is to determine the coefficients \( a \), \( b \), and \( c \) such that the given conditions are satisfied.

Setting Up the Problem (a)

For part (a), you need to substitute the values into \( p(x) = ax^2 + bx + c \):1. \( p(0) = 2 \) yields \( c = 2 \).2. \( p(1) = 3 \) results in \( a(1)^2 + b(1) + c = 3 \).3. \( p(3) = 8 \) results in \( a(3)^2 + b(3) + c = 8 \).Given \( c = 2 \), substitute into the equations for \( x = 1 \) and \( x = 3 \):- \( a + b + 2 = 3 \) simplifies to \( a + b = 1 \).- \( 9a + 3b + 2 = 8 \) simplifies to \( 9a + 3b = 6 \).

Solving the System of Equations (a)

Solve the system of equations obtained:1. \( a + b = 1 \)2. \( 9a + 3b = 6 \)To solve, first simplify equation 2 by dividing the whole equation by 3:- Equation 2 becomes \( 3a + b = 2 \).Now use substitution or elimination:- Subtract the first equation from the simplified second: \( (3a + b) - (a + b) = 2 - 1 \) simplifies to \( 2a = 1 \), giving \( a = \frac{1}{2} \).- Substitute \( a = \frac{1}{2} \) back into \( a + b = 1 \): \( \frac{1}{2} + b = 1 \), thus \( b = \frac{1}{2} \).The polynomial is \( p(x) = \frac{1}{2}x^2 + \frac{1}{2}x + 2 \).

Setting Up the Problem (b)

For part (b), set up the equations with \( p(x) = ax^2 + bx + c \):1. \( p(0) = 5 \) gives \( c = 5 \).2. \( p(1) = 3 \) becomes \( a(1)^2 + b(1) + 5 = 3 \).3. \( p(2) = 5 \) becomes \( a(2)^2 + b(2) + 5 = 5 \).Given \( c = 5 \), substitute into the equations for \( x = 1 \) and \( x = 2 \):- \( a + b + 5 = 3 \) simplifies to \( a + b = -2 \).- \( 4a + 2b + 5 = 5 \) simplifies to \( 4a + 2b = 0 \).

Solving the System of Equations (b)

Solve the system:1. \( a + b = -2 \)2. \( 4a + 2b = 0 \)First divide the second equation by 2 to simplify:- \( 2a + b = 0 \)Use substitution or elimination:- Subtract the second equation from the first: \( (a + b) - (2a + b) = -2 - 0 \) simplifies to \( -a = -2 \), hence \( a = 2 \).- Substitute \( a = 2 \) back in:\ \( 2(2) + b = 0 \), which gives \( 4 + b = 0 \), so \( b = -4 \).The polynomial is \( p(x) = 2x^2 - 4x + 5 \).

Key concepts

Polynomial Functions

Polynomial functions are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. The degree of a polynomial is the highest power of the variable in the expression. For example, a quadratic polynomial has a degree of 2, which typically appears in the form \( p(x) = ax^2 + bx + c \). It is characterized by its distinctive U-shaped graph called a parabola. This form allows it to model a wide range of phenomena, such as projectile motion or profit models. Quadratic polynomials can have constant, linear, and quadratic terms, with the quadratic term \( ax^2 \) dictating the curvature of the graph.

Understanding polynomial functions includes recognizing the importance of the degree of the polynomial and how each component impacts the overall function. Changing the coefficients affects both the shape and position of the parabola on a graph. A greater absolute value of \( a \) will make the parabola steeper, while different signs of \( a \) determine whether it opens upwards or downwards. The linear term, \( bx \), tilts the axis of symmetry, while the constant term \( c \) moves the graph up or down.

System of Equations

In solving problems relating to polynomial functions, systems of equations are often applied. This involves finding values for unknowns, like the coefficients \( a \), \( b \), and \( c \) in a polynomial, which satisfy multiple conditions simultaneously. The initial approach consists of forming equations based on problem constraints, often derived from given values of the polynomial for specific \( x \) values, like \( p(0) = 2 \), \( p(1) = 3 \), etc.

To solve a system of equations, methods such as substitution, elimination, or using matrices can be utilized. For instance, if you have the system:

• \( a + b = 1 \)
• \( 9a + 3b = 6 \)

you might simplify these equations for easier handling. Simplifying makes it possible to reduce one equation by systematic elimination of a variable, often leading to a single-variable equation. Solving this provides values that can be substituted back into other equations to find the remaining unknowns.

Coefficients of a Polynomial

Coefficients in a polynomial are scalar values that multiply each term of the polynomial. In the quadratic format \( ax^2 + bx + c \), \( a \), \( b \), and \( c \) are the coefficients, with \( c \) being the constant term. A polynomial's coefficients are crucial because they determine the function's shape and properties. They can often be discovered by substituting known values into the polynomial and forming equations.

The process typically involves plugging in values given in a problem to solve for coefficients. When a polynomial is fully described by points \((x_1, y_1), (x_2, y_2), \text{and} (x_3, y_3)\), one sets these values into equations:

• \( ax^2_1 + bx_1 + c = y_1 \)
• \( ax^2_2 + bx_2 + c = y_2 \)
• \( ax^2_3 + bx_3 + c = y_3 \)

These equations are then solved simultaneously, often requiring algebraic manipulation, to uncover precise coefficients. Adjustments in these coefficients will change the various properties, such as the axis of symmetry, vertex location, and intercepts of the polynomial's graph.