Problem 38 - Calculus Single Variable

Question

Find a polynomial function of degree 2 such that \(p(3)=4\) \(p(5)=-10,\) and \(p^{\prime}(-3)=7\).

Step-by-Step Solution

Answer

The polynomial is \( p(x) = -7x^2 + 49x - 80 \).

Step 1: Understand the Polynomial Form

A polynomial function of degree 2 is generally expressed as \( p(x) = ax^2 + bx + c \). We are tasked with finding the coefficients \( a \), \( b \), and \( c \) so that certain conditions are met.

Step 2: Set Up the Equations from Conditions

We have three conditions given: \( p(3) = 4 \), \( p(5) = -10 \), and \( p'(x) = 2ax + b \) with \( p'(-3) = 7 \). We can start by creating equations:1. \( a(3)^2 + b(3) + c = 4 \)2. \( a(5)^2 + b(5) + c = -10 \)3. \( 6a + b = 7 \).These simplify to:- \( 9a + 3b + c = 4 \)- \( 25a + 5b + c = -10 \)- \( 6a + b = 7 \).

Step 3: Solve for One of the Variables

Use the derivative condition \( 6a + b = 7 \) to solve for \( b \) in terms of \( a \):- \( b = 7 - 6a \).

Step 4: Substitute and Simplify the System

Substitute \( b = 7 - 6a \) into the other two equations:- From \( 9a + 3b + c = 4 \): \( 9a + 3(7 - 6a) + c = 4 \) \( 9a + 21 - 18a + c = 4 \) \( -9a + c = -17 \), giving \( c = 9a - 17 \).- From \( 25a + 5b + c = -10 \): \( 25a + 5(7 - 6a) + c = -10 \) \( 25a + 35 - 30a + c = -10 \) \( -5a + c = -45 \), giving \( c = 5a - 45 \).

Step 5: Equate and Solve for \( a \)

Since we have two expressions for \( c \), equate them to find \( a \):- \( 9a - 17 = 5a - 45 \) \( 4a = -28 \) \( a = -7 \).

Step 6: Find \( b \) and \( c \) Using \( a \)

Substitute back to find \( b \, \text{and} \, c \):- \( b = 7 - 6(-7) = 7 + 42 = 49 \)- \( c = 9(-7) - 17 = -63 - 17 = -80 \).

Step 7: Write the Polynomial Function

Now that we have \( a = -7 \), \( b = 49 \), and \( c = -80 \), the polynomial function is:\[ p(x) = -7x^2 + 49x - 80 \].

Key Concepts

Degree 2 Polynomials

Polynomials come in various degrees, which indicate the highest exponent of the variable in the expression. A degree 2 polynomial is often called a quadratic polynomial. The general form is

• \( p(x) = ax^2 + bx + c \)

where \( a \), \( b \), and \( c \) are coefficients, and \( a eq 0 \). This form represents a parabola when graphed. The presence of \( ax^2 \) as the term with the highest power is what classifies it as a degree 2 polynomial.
Quadratics are ubiquitous in mathematics due to their simple form that still allows for a rich variety of properties and solutions. When solving problems involving these polynomials, your primary goal is often to find the values of \( a \), \( b \), and \( c \) that satisfy specific conditions.
In our given exercise, you're tasked with determining these coefficients such that certain values of \( x \) yield given results. That's typical for many real-world applications, such as physics or engineering.

Derivative Conditions

Derivative conditions are crucial when working with polynomials, especially quadratics. The derivative of a polynomial function gives you information about its slope, or the rate of change, at any given point. For our degree 2 polynomial

• \( p(x) = ax^2 + bx + c \),

the derivative, denoted by \( p'(x) \), helps pinpoint turning points and slopes:

• \( p'(x) = 2ax + b \).

This first derivative simplifies the process of finding instantaneous rates of change and analyzing the curvature of the polynomial. In our context, one condition given is \( p'(-3) = 7 \). This means that at \( x = -3 \), the slope or the gradient of the tangent to the curve is 7.
Such conditions are invaluable for determining the specific values of \( a \) and \( b \) because they help you understand how the function behaves locally. By understanding the slopes, we can make more informed deductions about the overall shape of the curve.

Solving Systems of Equations

Solving systems of equations is a common task when dealing with polynomial problems. Often, multiple conditions require one to solve for several unknowns simultaneously. The essence of the process is to express all given conditions in terms of equations and then apply algebraic methods to find the unknown values.
In the exercise, you have three conditions provided as equations:

• \( p(3) = 4 \)
• \( p(5) = -10 \)
• \( p'(-3) = 7 \)

These lead to a system of three equations relating \( a \), \( b \), and \( c \). By substituting and solving these equations systematically, you can deduce the solution. For instance, one technique involves expressing one variable in terms of the others and substituting these expressions back into other equations. This often simplifies the system enough to find individual values for each variable.
In this problem, solving these equations results in the coefficients \( a = -7 \), \( b = 49 \), and \( c = -80 \). Solving systems empowers you to handle more complex equations based on multiple criteria. This skill is foundational in mathematics, ensuring all conditions are satisfied to find a precise solution.