Problem 83 - Precalculus

Question

Find \(b\) and \(c\) such that the parabola \(y=b x^{2}+b x+c\) goes through the points (4,46) and (-2,10).

Step-by-Step Solution

Answer

The values are \( b = 2 \) and \( c = 6 \).

Step 1: Substituting the first point into the equation

We start with the equation of the parabola: \( y = b x^2 + bx + c \). Since the parabola passes through the point (4, 46), substitute \( x = 4 \) and \( y = 46 \) into the equation: \[ 46 = b(4)^2 + b(4) + c \].This simplifies to: \[ 46 = 16b + 4b + c \] or \[ 46 = 20b + c \]. \quad (1)"

Step 2: Substituting the second point into the equation

Now take the second point (-2, 10) that the parabola passes through. Substitute \( x = -2 \) and \( y = 10 \) into the equation:\[ 10 = b(-2)^2 + b(-2) + c \].This simplifies to:\[ 10 = 4b - 2b + c \] or \[ 10 = 2b + c \]. \quad (2)"

Step 3: Solving the system of equations

Now, we have the following system of linear equations obtained from both points:1. \( 46 = 20b + c \)2. \( 10 = 2b + c \)Subtract the second equation from the first to eliminate \( c \):\[ (46 - 10) = (20b + c) - (2b + c) \]This simplifies to:\[ 36 = 18b \].Therefore:\[ b = 2 \]."

Step 4: Finding \( c \) using \( b \)

Now that we know \( b = 2 \), substitute this value back into one of the original equations to find \( c \). Using equation (2):\[ 10 = 2(2) + c \]\[ 10 = 4 + c \]Subtract 4 from both sides:\[ c = 6 \]."

Key Concepts

Solving Systems of Equations

Solving systems of equations means finding values for variables that satisfy all given equations simultaneously. These values make each equation true at the same time. In the context of a parabola, we often need information from multiple points to find the exact shape of the curve.

For example, when given two points that lie on the same parabola, this information provides us with a set of equations. Each point gives one equation when substituted into the parabolic formula, making up the 'system of equations'.

In our example, we derived:

• Equation 1: \( 46 = 20b + c \)
• Equation 2: \( 10 = 2b + c \)

The goal is to find the values of \( b \) and \( c \) that solve both equations at once. This requires methods like substitution or elimination, where we manipulate and combine the equations strategically.

Substitution Method

The substitution method is a handy technique for solving systems of equations. It involves solving one equation for one variable and then inserting or 'substituting' this expression into another equation.

In our parabola problem, after simplifying both points into equations, we have the system:

• \( 46 = 20b + c \)
• \( 10 = 2b + c \)

One effective way to use substitution here is to express \( c \) from either equation and plug it into the other equation. For example, we can subtract the second equation from the first to eliminate \( c \) directly, leading to an equation that only includes \( b \).

Once \( b \) is found, substituting it back into one of the original equations will produce the value of \( c \). This clear path ensures you uncover both unknown variables, completing the system of equations successfully.

Quadratic Functions

Quadratic functions are vital in the study of algebra and represent a foundational concept in understanding more complex mathematical ideas. A quadratic function generally takes the form \( y = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \).

These functions graph as parabolas, which could open upwards or downwards depending on the sign of \( a \). The parameters \( b \) and \( c \) impact the position and orientation of the parabola. For example, in the parabola equation \( y = b x^2 + b x + c \), the values of \( b \) and \( c \) must be determined accurately from the given points to adequately describe the function.

Quadratic functions often appear in real-world situations, such as calculating projectile motion or modeling scenarios with curved paths. Understanding how to derive these equations from points on the parabola allows you to predict and analyze countless situations effectively.