Problem 34 - Algebra and Trigonometry with Analytic Geometry

Question

If \(f(x)=a x^{3}+b x+c\), determine \(a, b\), and \(c\) such that the graph of \(f\) passes through the points \(P(-3,-12)\), \(Q(-1,22)\), and \(R(2,13)\)

Step-by-Step Solution

Answer

The coefficients are \( a = 2 \), \( b = -9 \), and \( c = 15 \).

Step 1: Identifying the Known Variables from Points

To solve for the coefficients of the cubic polynomial, we will substitute the known points given into the function. This will create a system of equations that can be solved for the unknowns.The function is given by: \( f(x) = ax^3 + bx + c \). Substitute each point into the function:- For point \( P(-3, -12) \): \(-12 = a(-3)^3 + b(-3) + c \)- For point \( Q(-1, 22) \): \( 22 = a(-1)^3 + b(-1) + c \)- For point \( R(2, 13) \): \( 13 = a(2)^3 + b(2) + c \).

Step 2: Substituting and Simplifying Equations

Substitute values for each point and simplify each expression:- \( -12 = a(-27) - 3b + c \) becomes \(-27a - 3b + c = -12 \)- \( 22 = a(-1) - b + c \) becomes \(-a - b + c = 22 \)- \( 13 = a(8) + 2b + c \) becomes \( 8a + 2b + c = 13 \).

Step 3: Setting Up the System of Equations

You now have a system of three linear equations:1. \( -27a - 3b + c = -12 \)2. \( -a - b + c = 22 \)3. \( 8a + 2b + c = 13 \).

Step 4: Solving the System of Equations

To find \( a \), \( b \), and \( c \), we can use different methods like substitution or elimination.1. Subtract the second equation from the first: \[ (-27a - 3b + c) - (-a - b + c) = -12 - 22 \] \[ -26a - 2b = -34 \] (Equation 4)2. Subtract the second equation from the third: \[ (8a + 2b + c) - (-a - b + c) = 13 - 22 \] \[ 9a + 3b = -9 \] (Equation 5)3. Simplify Equation 5: \[ 3a + b = -3 \] (Equation 6)4. Now solve Equations 4 and 6: Multiply Equation 6 by 2: \[ 6a + 2b = -6 \]5. Add Equation 4 and the modified Equation 6: \[ -26a - 2b + 6a + 2b = -34 - 6 \] \[ -20a = -40 \] \[ a = 2 \].

Step 5: Substituting Back to Find the Other Coefficients

Now that we have \( a = 2 \), substitute it back into the simplified Equation 6: \[ 3(2) + b = -3 \] \[ 6 + b = -3 \] \[ b = -9 \].Next, substitute \( a = 2 \) and \( b = -9 \) back into one of the original equations to find \( c \), let's choose Equation 2: \[ -2 - (-9) + c = 22 \] \[ 7 + c = 22 \] \[ c = 15 \].

Step 6: Conclusion

The coefficients that make the polynomial function pass through the given points are \( a = 2 \), \( b = -9 \), and \( c = 15 \). Thus, the function is \( f(x) = 2x^3 - 9x + 15 \).

Key Concepts

Cubic Equations

Cubic equations are polynomial equations of degree three. They have the standard form \( ax^3 + bx^2 + cx + d = 0 \). In these equations, \( a \) cannot be zero, or the equation would become quadratic, not cubic. The leading term is the term with the highest power, \( ax^3 \), which gives it its cubic nature. These functions can have a variety of graphs which might include one or two turning points, and can intersect the x-axis in up to three places.
The specific equation in the problem, \( f(x) = ax^3 + bx + c \), is a simplified version that omits the quadratic term \( bx^2 \) usually. This occurs sometimes based on specific properties of the function or additional conditions. Three coefficients \( a, b, \) and \( c \) influence the shape and position of the graph. Finding these coefficients is essential because they define how the curve passes through specific given points on a graph. This process requires creating and solving a system of equations, as demonstrated above.

System of Equations

A system of equations is essentially a set of two or more equations that share the same variables. These systems must be solved together to find a common solution for the variables. In this problem, we encounter a system of three equations since the polynomial must pass through three given points, forming three conditions. Each equation in the system corresponds to one point, with the variables being the coefficients \( a, b, \) and \( c \).
The importance of setting up your system of equations correctly cannot be overstated. It's the groundwork for solving more complex algebraic problems. Ensuring each equation accurately represents its respective condition will ensure that the system can be solved. Using methods like substitution and elimination can help to efficiently work through solving these equations. A correct solution to the system will yield the values for the coefficients needed to define the polynomial.

Solving Linear Equations

Solving linear equations within the system helps us find the values of \( a \), \( b \), and \( c \) in our polynomial. Linear equations, unlike cubics, have only one degree (the highest power of the variables is one). They take on the form \( ax + b = c \).
For our specific problem involving the cubic function, the strategy involves simplifying and rearranging the equations to isolate one of the variables. This could involve:

• Eliminating a specific variable by addition or subtraction of equations
• Substituting expressions back into other equations to simplify further
• Sequentially solving for each unknown through substitution
  

Methods like substitution and elimination are handy for deriving the required coefficients. Once a coefficient (like \( a = 2 \)) is found, it's easier to solve for the other coefficients using the simpler equations derived during the process.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and operations. They form the building blocks of algebra, providing the structure required to solve equations and model real-life scenarios mathematically.
For our cubic equation \( f(x) = ax^3 + bx + c \), each term is an algebraic expression. Here:

• \( ax^3 \) represents the cubic term, responsible for the shape of the graph
• \( bx \) is the linear component, influencing the slope
• \( c \) is the constant term, affecting the vertical shift of the polynomial function
  

Understanding each part enables students to grasp how altering coefficients affects the function's graph. Practically, manipulating these expressions by substituting values or rearranging them in equations allows us to solve for unknowns systematically. This process extends beyond this specific problem, introducing fundamental algebra skills necessary for tackling diverse mathematical concepts.