Problem 68 - Precalculus with Limits

Question

In Exercises 57-70, find any points of intersection of the graphs algebraically and then verify using a graphing utility. \(x^2+2y^2-4x+6y-5=0\) \(x^2-4x-y+4=0\)

Step-by-Step Solution

Answer

The points of intersection of the graphs are \((2.4, 0.4)\) and \((1, 1)\)

Step 1: Rearrange Equation

Rewrite the second equation in terms of y. From \(x^2-4x-y+4 = 0\), y can be isolated as \(y = x^2-4x+4\).

Step 2: Substituting the Value of y

Substitute \(y = x^2-4x+4\) into the first equation \(x^2 + 2y^2 - 4x + 6y - 5 = 0\). This results in a new equation: \(x^2 + 2(x^2 - 4x + 4)^2 - 4x + 6(x^2 - 4x + 4) - 5 = 0\).

Step 3: Solving for x

Solve the resulting quadratic equation for x. Simplify to get \(5x^2 - 20x + 12 = 0\). Solving this equation yields two solutions: \(x = 2.4\) and \(x = 1\).

Step 4: Solving for y

Substitute the values of x into the equation \(y = x^2 - 4x + 4\). When \(x = 2.4\), \(y = 0.4\). When \(x = 1\), \(y = 1\).

Step 5: Verification

The intersection points of the two given curves are \((2.4, 0.4)\) and \((1, 1)\). Verify these points visually or using a graphing utility by plotting the given equations and checking if they intersect at these points. The graphical view will confirm that the calculated results are correct.

Key Concepts

Understanding Quadratic Equations

Quadratic equations are a fundamental part of algebra and are crucial for solving many types of problems. A quadratic equation is generally written in the form \( ax^2 + bx + c = 0 \).
It involves a variable, usually represented as \(x\), raised to the second power. These equations often form a parabola when graphed.

• Coefficient \(a\): Determines the opening direction of the parabola.
• Coefficient \(b\): Influences the slope and symmetry of the parabola.
• Constant \(c\): Represents the y-intercept of the graph.

When solving the intersection of two quadratic equations, you generally solve each equation separately and then find common solutions. It’s all about finding where two parabolas cross each other on a graph.

The Algebraic Solution Process

The algebraic method is all about substituting and rearranging equations to find solutions. In the given problem, we turn to algebra to find the point where two different equations intersect. This is done by:

• Rearranging the second equation to isolate \(y\).
• Substituting this expression for \(y\) into the first equation.

This substitution transforms the system into a single quadratic equation, which can be solved using techniques such as factoring, applying the quadratic formula, or simplifying the expressions.
After simplifying, we may end up with a manageable quadratic equation like \(5x^2 - 20x + 12 = 0\).Solving this gives us the x-coordinates of the intersection points. Finally, use these x values to find corresponding y values by substituting back into the rearranged equation.

Using a Graphing Utility

Once the algebraic solution is obtained, verifying it with a graphing utility is a smart move to ensure accuracy. Graphing utilities can be physical calculators or software applications like Desmos or GeoGebra. They allow you to plot equations and visually inspect intersection points. Here's how you use them:

• Input each equation into the graphing software.
• Observe where the graphs intersect visually.

These visual tools make it easy to confirm the algebraic solutions are correct. They also help you understand the spatial relationship between curves. This ability to visualize the entire graph is extremely useful, especially when dealing with complex equations and inequalities.

Substitution Method Simplified

The substitution method is a classic technique used for solving systems of equations. The goal is to eliminate one variable by expressing it in terms of another. In our instance:

• Start with identifying one equation to express one variable in terms of the other.
• Substitute that expression into the other equation.

This turns a system of two equations into a single equation with just one variable. The remaining equation can then be solved using standard techniques.
For the intersection problem:

• Expression for \(y\) was derived from the second equation.
• This expression was substituted into the first equation to solve for \(x\).

After finding the \(x\) values, substitute back to find \(y\). This structured approach simplifies solving complex algebraic problems and is very effective for consistent results.