Problem 31 - The Calculus with Analytic Geometry

Question

Draw a sketch of the graph of each of the following equations: (a) \(x+3 y=0\) (b) \(x-3 y=0\) (c) \(x^{2}-9 y^{2}=0\)

Step-by-Step Solution

Answer

The graphs for (a) and (b) are straight lines through the origin with slopes of -1/3 and 1/3 respectively. Graph (c) is a hyperbola formed by the intersecting lines x = 3y and x = -3y.

Step 1: Solve for y in the equation (a): x + 3y = 0

Rearrange the equation to isolate y. \[ x + 3y = 0 \]Subtract x from both sides: \[ 3y = -x \]Divide by 3: \[ y = -\frac{x}{3} \]

Step 2: Plot the equation (a): y = -x/3

This is a linear equation with a slope of -1/3 and a y-intercept of 0. Draw a straight line that passes through the origin (0,0) with a negative slope. The line goes downwards as it moves to the right.

Step 3: Solve for y in the equation (b): x - 3y = 0

Rearrange the equation to isolate y. \[ x - 3y = 0 \]Subtract x from both sides: \[ -3y = -x \]Divide by -3: \[ y = \frac{x}{3} \]

Step 4: Plot the equation (b): y = x/3

This is a linear equation with a slope of 1/3 and a y-intercept of 0. Draw a straight line that passes through the origin (0,0) with a positive slope. The line goes upwards as it moves to the right.

Step 5: Rewrite the equation (c): x^2 - 9y^2 = 0

Factor the equation. \[ x^2 - 9y^2 = 0 \]This can be written as a difference of squares: \[ (x - 3y)(x + 3y) = 0 \]

Step 6: Solve for x

Set each factor equal to 0 and solve for x. \[ x - 3y = 0 \]\[ x = 3y \]and \[ x + 3y = 0 \]\[ x = -3y \]

Step 7: Plot the hyperbola

The equations \( x = 3y \) and \( x = -3y \) are the equations of two intersecting lines forming an 'X'. Plot these as lines passing through the origin with slopes of 3 and -3 respectively.

Key Concepts

linear equations

Linear equations are equations of the first degree, meaning they graph as straight lines. These equations typically take the form \[ ax + by = c \], where \(a\), \(b\), and \(c\) are constants. The key to solving linear equations is isolating the variable, commonly \(y\) or \(x\). For example, in the equation \(x + 3y = 0\), isolate \(y\) by subtracting \(x\) from both sides and then dividing by 3. After solving, you can identify important features such as the slope and y-intercept to plot the graph efficiently.

slope-intercept form

The slope-intercept form of a linear equation is \(y = mx + b\), where \(m\) represents the slope and \(b\) is the y-intercept. Knowing how to convert equations to this form is crucial for graph plotting. For instance, the equation \(x + 3y = 0\) can be rearranged to \(y = -\frac{x}{3}\), showing a slope \(m = -\frac{1}{3}\) and y-intercept \(b = 0\). In contrast, the equation \(x - 3y = 0\), when rearranged, becomes \(y = \frac{x}{3}\), highlighting a slope \(m = \frac{1}{3}\).

hyperbola

A hyperbola is a type of conic section represented by equations such as \(x^2 - 9y^2 = 0\). This specific equation can be factored into \((x - 3y)(x + 3y) = 0\), indicating it separates into two intersecting lines rather than a conventional 'hyperbola' shape. Solving gives us \(x = 3y\) and \(x = -3y\), two linear equations that form an 'X' on the graph. Hyperbolas typically involve asymptotes, but in this case, these lines represent the critical intersecting points of the hyperbola-like shape.

solving equations

Solving equations involves finding the values that satisfy the equation. For linear equations like \(x + 3y = 0\), solve for one variable by isolating it with algebraic operations. Factors such as slopes and intercepts become essential once equations are solved. For example, \(x = 3y\) and \(x = -3y\), derived from the hyperbola equation, are solved by setting each factor to zero. This systematic approach helps in converting complex equations into simpler, solvable parts.

plotting graphs

Plotting graphs effectively illustrates solutions visually. Start by identifying the type of equation: linear, quadratic, hyperbola, etc. For linear equations \(y = -\frac{x}{3}\) and \(y = \frac{x}{3}\), plot points or use the slope-intercept form to draw lines. The intersection points and slopes guide accurate placement. With hyperbolas like \(x^2 - 9y^2 = 0\), the derived linear equations \(x = 3y\) and \(x = -3y\) are plotted as lines that intersect. Visual representation through graphs reinforces understanding and verifies solution accuracy.