Problem 42 - Contemporary Precalculus

Question

Find the \(x\) - and \(y\) -intercepts of the graph of the equation. $$x^{2}-2 x y+3 y^{2}=1$$ CAN'T COPY THE GRAPH

Step-by-Step Solution

Answer

Answer: The x-intercepts are \(x=1, -1\), and the y-intercepts are \(y = \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\).

Step 1: Find the x-intercepts

Set y=0 in the given equation: $$x^2 - 2x(0) + 3(0)^2 = 1$$ Simplify the equation to find the x-intercepts: $$x^2 = 1$$

Step 2: Solve for x

Use the square root property: $$x = \pm\sqrt{1}$$ This gives us two x-intercepts: $$x = 1, -1$$

Step 3: Find the y-intercepts

Set x=0 in the given equation: $$(0)^2 - 2(0)y + 3y^2 = 1$$ Simplify the equation to find the y-intercepts: $$3y^2 = 1$$

Step 4: Solve for y

Divide by 3 and use the square root property: $$y^2 = \frac{1}{3}$$ $$y = \pm\sqrt{\frac{1}{3}}$$ This gives us two y-intercepts: $$y = \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}$$

Step 5: Write the final answer

The x-intercepts are \(x=1, -1\), and the y-intercepts are \(y = \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\).

Key Concepts

Quadratic Equations

Quadratic equations are a common feature in algebra and represent a parabolic graph when plotted. They have the general form \( ax^2 + bx + c = 0 \), where \( a \), \( b \) and \( c \) are constants, and \( a \) is not zero. This specific form enables a wide range of methods to be applied to solve for the variable \( x \).

In our exercise, the equation \( x^2 - 2xy + 3y^2 = 1 \) appears to be a quadratic equation in two variables \( x \) and \( y \). To find the \( x \) and \( y \) intercepts, one often sets the other variable to zero and solves the resulting equation. When \( y=0 \), we get a simple quadratic equation in \( x \), and when \( x=0 \) we get a quadratic equation in \( y \). Solving these results in the points where the graph intersects the respective axes.

Square Root Property

The square root property is particularly useful in solving quadratic equations when they are in their simplest form \(x^2 = k\), where \(k\) is any real number. Using this property, we can say that \(x = ±\text{sqrt}(k)\), which means the solutions are the principal square root and its negative counterpart.

In the exercise, after setting one variable to zero, we were able to simplify the equation to this form, either \( x^2 = 1 \) or \( y^2 = \frac{1}{3} \). By applying the square root property, we find the possible values of \( x \) and \( y \) that satisfy the equation. For example, the application of the square root property to \( x^2 = 1 \) results in two solutions, \(x = 1\) and \(x = -1\), which are the \( x \) intercepts of the graph.

Graphing Equations

Graphing equations is a visual way of representing the solutions to an equation. It involves drawing the equation on the Cartesian coordinate system, where the \( x \) and \( y \) intercepts are particularly important as they define where the graph crosses the axes. These intercepts are often the starting points for graphing.

In the given problem, the goal was to find both the \( x \) and \( y \) intercepts for the equation \( x^2 - 2x y + 3y^2 = 1 \). Once we obtained these intercepts through the algebraic process, they could be plotted on the graph as points \( (1,0), (-1,0), (0, \frac{1}{\text{sqrt}(3)}), \) and \( (0, -\frac{1}{\text{sqrt}(3)}) \). These points are essential for sketching the shape and position of the parabola or other graphs that might result from the original equation.