Problem 21 - Algebra and Trigonometry

Question

Two equations and their graphs are given. Find the intersection point(s) of the graphs by solving the system. \(\left\\{\begin{aligned} x^{2}+y &=0 \\ x^{3}-2 x-y &=0 \end{aligned}\right.\)

Step-by-Step Solution

Answer

The graphs intersect at points (0, 0), (1, -1), and (-2, -4).

Step 1: Express y from the First Equation

The first equation in the system is \(x^2 + y = 0\). We solve for \(y\) in terms of \(x\): \(y = -x^2\).

Step 2: Substitute y into the Second Equation

Now that we have \(y = -x^2\), substitute this expression into the second equation \(x^3 - 2x - y = 0\), resulting in \(x^3 - 2x - (-x^2) = 0\). Simplifying gives us \(x^3 - 2x + x^2 = 0\).

Step 3: Rearrange and Simplify the New Equation

The equation from Step 2 is \(x^3 + x^2 - 2x = 0\). Factor out an \(x\) to simplify: \(x(x^2 + x - 2) = 0\).

Step 4: Factor the Quadratic

Next, factor the quadratic \(x^2 + x - 2\). This can be factored as \((x - 1)(x + 2)\). So now the equation becomes \(x(x - 1)(x + 2) = 0\).

Step 5: Solve for x

We solve \(x(x - 1)(x + 2) = 0\) to find the values of \(x\). This gives us three solutions: \(x = 0\), \(x = 1\), and \(x = -2\).

Step 6: Find Corresponding y-values

For each value of \(x\), substitute back into \(y = -x^2\) to find the corresponding \(y\):- If \(x = 0\), then \(y = 0\).- If \(x = 1\), then \(y = -1\).- If \(x = -2\), then \(y = -4\).

Step 7: Identify the Intersection Points

From the previous steps, the intersection points where the graphs meet are: \((0, 0)\), \((1, -1)\), and \((-2, -4)\).

Key Concepts

Intersection Points in a System of Equations

When you have two equations and their graphs, the intersection points are the locations where these graphs meet on a coordinate plane. Finding these points is crucial because they represent solutions that satisfy both equations simultaneously.
To find these points, we solve the given system of equations. For the system provided, decomposition of the problem into smaller ones helps in identifying intersections effectively.

• The first step is to express one variable in terms of the other using one of the equations.
• Next, substitute this expression into the other equation to find the potential intersection points.

The process ends when respective values of x and y that satisfy both equations are identified. In this example, the solutions provide the three intersection points:

• \((0, 0)\) - where both equations are simultaneously satisfied when \(x = 0\).
• \((1, -1)\) - another solution that fits both conditions.
• \((-2, -4)\) - the final intersection detected.

Understanding Graphing with Systems of Equations

Graphing is an excellent visual method for solving systems of equations. It provides a clear depiction of where the equations intersect, which are solutions to the system.
Graphing involves creating the graphical representations of each equation by plotting points or using transformations. Once graphed, the solution points or intersections can be visually identified.
For non-linear equations like polynomials, graphing is particularly useful. Curved graphs can intersect in multiple places or might not intersect at all, giving a more comprehensive picture of solutions.
In the problem given, if you graph

• the equation \( y = -x^2 \)
• the equation \( y = x^3 - 2x \)

it becomes much easier to see where the intersections take place. By graphing these equations, the intersection points can be visually confirmed as exactly the spots where both graphs touch each other.

Solving Systems with Algebraic Solutions

An algebraic solution involves manipulating equations to isolate variables and deduce the solutions of a system precisely. This method is essential for exact solutions without needing graphical approximations.
When you're solving systems algebraically, especially with non-linear systems such as the one in our exercise, the following steps are practical:

• First, express one variable in terms of the other using one of the given equations. In this example, solving for \(y\) yields \(y = -x^2\).
• Next, substitute this expression in the other equation, allowing us to create a single equation involving just one variable.

For the given example:

• Substitute \(y = -x^2\) into \(x^3 - 2x - y = 0\).
• Solve the resulting equation \(x^3 + x^2 - 2x = 0\).

This process reveals the exact values of \(x\) and subsequently the corresponding \(y\) values. The algebraic approach guarantees accurate intersection points and allows us to confidently claim the results without further verification.