Problem 25 - Algebra 2

Question

Use the equation \(A x^{2}+B x y+C y^{2}+D x+E y+F=0\) to identify the shape of the graph that results in each case. a. \(A=C=D=0, B \neq 0, F \neq 0\) b. \(A=B=C=0, D \neq 0, E \neq 0, F \neq 0\)

Step-by-Step Solution

Answer

For both cases a and b: The shape of the graph results in a line.

Step 1: Identify the shape in case a

Given the coefficients \(A=C=D=0, B \neq 0, F \neq 0\). Substituting these values in the equation \(A x^{2}+B x y+C y^{2}+D x+E y+F=0\), it simplifies to \(B x y + E y +F=0\). This equation is a linear equation in y, which means the shape of the graph will be a line.

Step 2: Identify the shape in case b

Given the coefficients \(A=B=C=0, D \neq 0, E \neq 0, F \neq 0\). Substituting these values in the equation \(A x^{2}+B x y+C y^{2}+D x+E y+F=0\), it simplifies to \(Dx + Ey + F = 0\). This is also a linear equation, hence the graph in this case will also be a line.

Key Concepts

Equations of Lines

An equation of a line is often written in the form of either slope-intercept form, point-slope form, or standard form. For instance, the standard form is given by \( Ax + By + C = 0 \). Here, \( A \), \( B \), and \( C \) are constants, and \( x \) and \( y \) are variables that represent coordinates on a two-dimensional graph.

In situations where values of \( A \), \( B \), and \( C \) change, the orientation and position of the line on the graph will also change. If we have a non-zero coefficient in front of either \( x \) or \( y \), it ensures that the equation will form a line rather than a different geometric figure. This is evident from the exercise, where the modified equations in both scenarios resulted in linear graphs.

Understanding these basics allows us to predict the shape of graphs based on their equations, particularly linear equations, which are the simplest form of polynomial equations.

Graph Shapes

Graphs of equations can take on various shapes, and the type of shape is largely dictated by the equation's degree and its coefficients. Line graphs represent a first-degree polynomial equation where terms are just linear combinations of \( x \) and \( y \).

From the exercise examples given:

• In case (a), the equation simplifies to a linear form \( Bxy + Ey + F = 0 \), which denotes a line in the xy-plane.
• Similarly, in case (b), the equation \( Dx + Ey + F = 0 \) also results in a line.

When you encounter an equation of a conic section, examining the coefficients quickly helps identify which type of conic section it represents. If all terms with \( x^2 \) or \( y^2 \) are zero, we generally end up with linear equations representing lines.

Recognizing these patterns helps students identify and graph the correct shape without necessarily solving the equation entirely each time.

Solving Quadratic Equations

Quadratic equations are those of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and at least the squared variable term is present. Solving quadratic equations can typically result in one of three outcomes for graph shapes: parabolas, ellipses, or hyperbolas.

Key solving methods include:

• Factoring: Useful when the equation neatly breaks down into products of binomials.
• Quadratic Formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) gives solutions for any solvable quadratic equation.
• Completing the Square: Alters the equation to a form that easily shows vertex for graphing parabolas.

In conic sections, it is crucial to simplify the equation to see if it reduces to a quadratic form. In the provided exercise, by checking coefficients, we skipped this step because when quadratic terms were absent, it directly indicated the lines as outputs.

Often, understanding the underlying geometry can aid in solving quadratic equations by realizing what kind of graph they depict on a coordinate plane.