Problem 86 - Precalculus

Question

The parabola \(y=b x^{2}-2 x-a\) goes through the points \((-2, a)\) and \((-1, b-2) .\) Find \(a\) and \(b\).

Step-by-Step Solution

Answer

The solutions are \(a = 4\) and \(b = 1\).

Step 1: Analyze the Given Equation and Points

The parabolic equation is given as \( y = b x^2 - 2x - a \), and it passes through the points \((-2, a)\) and \((-1, b-2)\). We will plug these points into the equation to set up two equations with variables \(a\) and \(b\).

Step 2: Substitute the First Point into the Parabola Equation

Substitute the point \((-2, a)\) into the parabola equation. This gives us:\[a = b(-2)^2 - 2(-2) - a\]Simplifying the expression, we get:\[a = 4b + 4 - a\]which results in:\[2a = 4b + 4\]Thus, we have Equation (1):\[a = 2b + 2\]

Step 3: Substitute the Second Point into the Parabola Equation

Now, substitute the point \((-1, b-2)\) into the parabola equation:\[b - 2 = b(-1)^2 - 2(-1) - a\]Simplifying, we have:\[b - 2 = b + 2 - a\]This simplifies to:\[a = 4\]Thus, we have Equation (2):\[a = 4\]

Step 4: Solve the System of Equations

Using Equation (2), \(a = 4\), substitute \(a\) into Equation (1):\[4 = 2b + 2\]Simplifying, we have:\[2b = 2\]Thus, solving for \(b\):\[b = 1\]Therefore, the values of \(a\) and \(b\) are \(a = 4\) and \(b = 1\).

Key Concepts

Quadratic Equations

A quadratic equation in its standard form is typically written as \(ax^2 + bx + c = 0\). However, equations can often be arranged differently, as we have in the exercise, where the quadratic is expressed as \(y = bx^2 - 2x - a\). Here, the term quadratic denotes the highest power of the variable, which is 2. This indicates the shape of the graphed equation will be a parabola—a symmetric curve.To understand a quadratic equation fully, it's important to grasp its components:

• The coefficient of \(x^2\) dictates the parabola's "width" and orientation. A positive value opens upwards, while a negative one opens downwards.
• The linear term, which here is \(-2x\), helps in determining the position of the vertex.
• The constant term, denoted as \(-a\), shifts the parabola up or down the y-axis.

The role of quadratic equations is vast—from calculating projectile paths in physics to modeling business profits.

Coordinate Geometry

Coordinate geometry involves graphing equations on a plane defined by an \(x\)-axis and \(y\)-axis. This exercise presents a parabolic curve residing in this coordinate plane, determined by the quadratic function listed.Each point on the parabola corresponds to an \(x\) and \(y\) pair that resolves the equation. In our case, the parabola passes through the points \((-2, a)\) and \((-1, b-2)\). By plugging these coordinates into the equation, we gain a system of mathematical statements. These help us find exact values for given variables.Visualizing these points on a graph assists in understanding how the parabola interacts with coordinate axes. Note, effective use of the coordinate geometry also helps determine the slope and curvatures of paths—crucial for tackling complex problems involving motion and optimization.

System of Equations

A system of equations is a collection of multiple equations that share variables. In our exercise, the challenge is to find the values of \(a\) and \(b\) using the information from two points on a parabolic curve.Here's how it works:

• Substitute each point into the quadratic equation to create separate equations with the same variables.
• Solve these equations simultaneously to find a common solution, representing where the equations intersect.

This method allows us to determine unique values for variables and assess relationships between variables in complex mathematical situations.Solving systems of equations is a powerful skill in mathematics, useful in anything from computational algorithms to physics and engineering.