Problem 20 - Algebra and Trigonometry

Question

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

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 solve for \(x\):\[x = y + 2\]

Step 2: Substitute x in the first equation

Substitute \(x = y + 2\) into the first equation \(x-y^2 = -4\):\[(y + 2) - y^2 = -4\]

Step 3: Simplify and solve the quadratic equation

Simplify the expression from Step 2:\[y + 2 - y^2 = -4\] Rearrange to form the quadratic equation:\[y^2 - y - 6 = 0\] Solve this quadratic equation using the quadratic formula, where \(a = 1\), \(b = -1\), and \(c = -6\):\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] This simplifies to: \[y = \frac{1 \pm \sqrt{1 + 24}}{2}\] \[y = \frac{1 \pm 5}{2}\] This gives two solutions: \(y = 3\) and \(y = -2\).

Step 4: Find corresponding x values

Substitute each \(y\) value back into \(x = y + 2\): - For \(y = 3\): \(x = 3 + 2 = 5\) - For \(y = -2\): \(x = -2 + 2 = 0\).

Step 5: Write the intersection points

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

Key Concepts

Intersection Points

The concept of intersection points is all about finding where two or more graphs meet on a coordinate plane. When graphs intersect, the points where they cross each other represent solutions that satisfy all equations involved.

• For example, in your exercise, two graphs meet at points, meaning that both equations are true for these values of \(x\) and \(y\).
• These points are critical because they represent the actual solutions to the system of equations.

Visualizing a graph or plotting the points can help you understand how the functions interact. In this exercise, solving the system algebraically gives you a precise answer without needing the graph but understanding intersections on a graph strengthens your concept.
Key takeaway: To find intersection points, solve the system to find \(x\) and \(y\) values that satisfy both equations together.

Quadratic Equation

A quadratic equation is an equation of the form \(ax^2 + bx + c = 0\). In your exercise, the quadratic appears when rearranging the substituted equation to solve for the variable \(y\).

• These equations have a characteristic 'U' shaped graph called a parabola. They can have zero, one, or two solutions depending on the discriminant \(b^2 - 4ac\).
• In the step-by-step solution, you saw how to convert a problem into a quadratic form and solve it using the quadratic formula.

The quadratic formula is \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), which helps you find the roots of any quadratic equation. Knowing how to rearrange and identify \(a\), \(b\), and \(c\) correctly is essential.
Key takeaway: Use the quadratic formula to solve when your equation fits \(ax^2 + bx + c = 0\), and it helps to understand its graphical meaning - intersecting the x-axis at real roots.

Substitution Method

The substitution method is a technique used to solve systems of equations by expressing one variable in terms of another and then substituting into another equation. This method effectively turns one of the equations into a one-variable equation.

• In your exercise, you first solve the second equation for \(x\): \(x = y + 2\).
• By substituting \(x\) with \(y + 2\) in the first equation, you create a single equation in one variable (in this case, \(y\)).

Substitution is especially powerful when one equation is easily solvable for a single variable, simplifying further calculations. Once you solve for the first variable, substitute back to find the second one.
Key takeaway: Use substitution to simplify systems of equations to one variable, making them easier to solve step by step.