Problem 16 - Algebra and Trigonometry

Question

\(15-18\) . Two equations and their graphs are given. Find the inter- section point(s) of the graphs by solving the system. $$ \left\\{\begin{array}{l}{x-y^{2}=-4} \\ {x-y=2}\end{array}\right. $$

Step-by-Step Solution

Answer

The intersection points are \((5, 3)\) and \((0, -2)\).

Step 1: Express x in terms of y from the second equation

From the equation \(x - y = 2\), we can express \(x\) in terms of \(y\) by adding \(y\) to both sides. This gives us: \(x = y + 2\).

Step 2: Substitute x in the first equation

Substitute the expression for \(x\) from Step 1 into the first equation \(x - y^2 = -4\). This gives us \(y + 2 - y^2 = -4\).

Step 3: Simplify the equation

Rearrange the equation from Step 2 to form a quadratic equation: \(-y^2 + y + 2 = -4\). Add 4 to both sides to get: \(-y^2 + y + 6 = 0\).

Step 4: Solve the quadratic equation

To solve \(-y^2 + y + 6 = 0\), multiply the entire equation by -1 to get \(y^2 - y - 6 = 0\). Factor the quadratic as \((y - 3)(y + 2) = 0\). The solutions for \(y\) are \(y = 3\) and \(y = -2\).

Step 5: Find corresponding x-values

Substitute the values of \(y\) back into \(x = y + 2\) to find \(x\). For \(y = 3\), \(x = 3 + 2 = 5\).For \(y = -2\), \(x = -2 + 2 = 0\).

Step 6: Write the intersection points

Based on the values found in Steps 4 and 5, the intersection points are \((5, 3)\) and \((0, -2)\).

Key Concepts

Quadratic Equations

Quadratic equations are equations that can be written in the form \(ax^2 + bx + c = 0\) where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). These equations form parabolas when graphed on a coordinate plane.To solve quadratic equations, we can use different methods such as factoring, using the quadratic formula, or completing the square. In our problem, we encountered a quadratic equation during the process of finding intersection points.

• Initially, we derived the equation \(-y^2 + y + 6 = 0\), then converted it to a standard form \(y^2 - y - 6 = 0\).
• This equation was factored into \((y - 3)(y + 2) = 0\) indicating two solutions for \(y\): \(y = 3\) and \(y = -2\).

Understanding how to work with quadratic equations is a fundamental skill in algebra. They frequently appear in a variety of mathematical contexts, including systems of equations.

Intersection Points

Intersection points are where two or more graphs meet each other on a coordinate plane. They represent the solution set for a system of equations. Finding these points can involve algebraic or graphical methods, and the solutions tell us the values of variables that satisfy all equations in the system simultaneously.For the given problem, we were tasked with finding intersection points of the equations:

• We solved the system algebraically by expressing \(x\) in terms of \(y\) (\(x = y + 2\)), and substituting it into another equation.
• The resulting solved values of \(y\) led us to the two points of intersection: \((5, 3)\) and \((0, -2)\).

Knowing where graphs intersect helps in understanding solutions to simultaneous equations, and it's a core idea in collegial algebra.

Graphical Solutions

Graphical solutions give a visual representation of equations on a coordinate plane. It offers an intuition-based approach where you can actually see how equations meet or differ.For our system, if we were to graphically represent the equations \(x - y = 2\) and \(x - y^2 = -4\), we would:

• Plot the linear equation \(x - y = 2\) as a straight line.
• Graph the quadratic equation \(x - y^2 = -4\) as a parabola.
• The intersection points of these graphs would appear where the line and parabola cross, which aligns with our algebraically determined solutions: \((5, 3)\) and \((0, -2)\).

Graphing can sometimes offer a more immediate understanding of solutions, making it an excellent tool for verifying algebraic results. This visual aid is particularly useful in identifying and confirming intersection points.