Analytic Geometry with Solid Mensuration - Algebraic Curves and Equations

Algebraic Curves and Equations

Competencies

• Determine the relations between curve and equation.

• Draw and trace the different curves by plotting of points.

• Identify the different properties of algebraic curves.

• Identify the different types of algebraic curves.

• Enumerate the steps in tracing the curves.

Relation Between Curve and Equation

• An equation involving variables $x$ and $y$ is typically satisfied by infinite pairs of values.

• Each pair corresponds to a point.

• These points form a geometric figure called a curve or locus.

• Each algebraic equation corresponds to a definite curve or locus or no locus at all.

Drawing a Curve by Point-Plotting

• A curve can be drawn by point-plotting: assign values to $x$ and solve for $y$, or vice versa.

• Each pair of values $(x, y)$ represents a point.

• Connect a sufficient number of points with a smooth curve to approximate the required curve.

• The number of points varies for different curves; more points yield a more accurate curve.

• Tabulating values of $x$ and $y$ can facilitate plotting.

Example:

• Plot the curves represented by the equations:

  • $y = 2x - 5$

  • $x^2 = 4y + 4$

Properties of Algebraic Curves

• Point-plotting can be long and tedious.

• Examining and interpreting certain properties can simplify determining the general appearance of a curve.

• Key properties include intercepts, symmetry, scope, and asymptotes.

Intercepts

• Intercepts are points where the curve crosses the coordinate axes.

• The $x$-intercepts are the abscissas of the points where the curve crosses the $x$-axis.

• The $y$-intercepts are the ordinates of the points where the curve crosses the $y$-axis.

• To find $x$-intercepts, set $y = 0$ and solve for $x$.

• To find $y$-intercepts, set $x = 0$ and solve for $y$.

Symmetry

• The center of symmetry of two points $A$ and $B$ is the midpoint $M$ between them.

• The axis (or line) of symmetry of two points is the perpendicular bisector of the line joining them.

• A curve is symmetric with respect to a coordinate axis if for every point on one side, there's a corresponding image point on the opposite side.

• A curve is symmetric with respect to the origin if for every point, there's a corresponding image point directly opposite and at an equal distance from the origin.

Tests for Symmetry

1. If an equation is unchanged by substituting $-y$ for $y$, the curve is symmetric with respect to the $x$-axis.

2. If an equation is unchanged by substituting $-x$ for $x$, the curve is symmetric with respect to the $y$-axis.

3. If an equation is unchanged by substituting $-x$ for $x$ and $-y$ for $y$ simultaneously, the curve is symmetric with respect to the origin.

Scope of the Curve

• Solve the equation for $y$ (or $y^2$) to determine if the curve lies above or below the $x$-axis in certain areas.

• Note the changes of sign.

• These areas are called regions, bounded by vertical lines through the intersections of the curve with the $x$-axis.

Asymptote

• A straight line is an asymptote of a curve if the curve approaches the line more and more closely but never really touches it, except as a limiting position at infinity.

• Not all curves have asymptotes.

• Asymptotes help in drawing the curve.

• Types of asymptotes include horizontal, vertical, oblique, and curvilinear.

Types of Algebraic Curves

• A polynomial is a mathematical expression with variables, coefficients, and operations of addition, subtraction, multiplication, and non-negative integer exponents.

• A polynomial equation is an equality between two polynomials.

• An algebraic curve is a curve defined by a polynomial equation in two variables, typically $x$ and $y$.

Examples:

• Polynomial: $5x^2 + 2y - 7$ (degree is 2)

• Polynomial equation: $2x^3 - 3y = -7x + y^2$

• Only curves whose equations can be written in the form polynomial in $x$ and $y = 0$ are considered algebraic curves.

• Curves whose equations cannot be expressed in this form are called transcendental curves and higher plane curves.

• Examples of transcendental curves are logarithmic and trigonometric functions.

• Cubic and quartic polynomial curves are examples of higher plane curves.

• Conic sections are examples of second-degree algebraic curves.

Types of Algebraic Curves based on equations

Let $f(x)$ and $g(x)$ be polynomials. Four types of algebraic curves can be defined as:

1. $y = f(x)$ (e.g., $y = 2x^2 + 3$)

2. $y = \frac{f(x)}{g(x)}$ (e.g., $y = \frac{2x^3 + x - 1}{7 - 2x^2}$

3. $y^2 = f(x)$ (e.g., $y^2 = 4x + 3$)

4. $y^2 = \frac{f(x)}{g(x)}$ (e.g., $y^2 = \frac{4x^3 - 1}{3 - 2x}$)

• Types 3 and 4 are symmetric with respect to the $x$-axis.

• Only types 2 and 4 have vertical and horizontal asymptotes.

• This course will focus on types 1 and 2.

Steps in Tracing a Curve

1. If the equation is given in the form $f(x, y) = 0$, solve for $y$ (or $y^2$) to express it in one of the four general types.

2. Test for symmetry by inspection using the simplified tests.

3. Determine the intersections with the $x$- and $y$-axes:

   • For $x$-intercepts, set $y = 0$, factor the right-hand side, and solve for $x$. This is the factored form of $y$.

   • For $y$-intercepts, set $x = 0$ and solve for $y$.

4. Determine the asymptotes, if any:

   • For vertical asymptotes, use the factored form. Let $y \rightarrow \pm \infty$, then solve for $x$. The line $x = h$ is a vertical asymptote.

   • For horizontal asymptotes, use the general form. Divide each term by the highest degree $x$-term. Then let $x \rightarrow \pm \infty$ and solve for $y$. The line $y = k$ is a horizontal asymptote.

   • Determine intersections of the curve with horizontal asymptotes by substituting $y = k$ in the equation and solving for $x$.

5. Divide the plane into regions using vertical lines through the intersections on the $x$-axis and vertical asymptotes. Don't consider the $y$-axis unless there is an intersection at the origin. Find the sign of $y$ in each region using the factored form to determine if the curve lies above or below the $x$-axis.

6. Trace the curve, plotting a few points if necessary.

Important Notes on Infinity

If $a$ is any real number and $\infty$ is a very large number:

1. $\frac{a}{0} = \infty$

2. $\frac{0}{a} = 0$

3. $\frac{a}{\infty} = 0$

The symbol $\infty$ represents infinity and is not a number.

Problem-Solving Examples

Example 1: Finding Intercepts

Find the intercepts of the curve $y = x^3 - 6x^2 + 3x + 10$.

Solution:

• X-intercepts:

  • Set $y = 0$: $x^3 - 6x^2 + 3x + 10 = 0$

  • Using the rational root theorem, possible roots are $\pm 10, \pm 5, \pm 2, \pm 1$

  • Let $f(x) = x^3 - 6x^2 + 3x + 10$

  • By trial and error:

    • $f(-1) = (-1)^3 - 6(-1)^2 + 3(-1) + 10 = -1 - 6 - 3 + 10 = 0 \Rightarrow x = -1$

    • $f(2) = (2)^3 - 6(2)^2 + 3(2) + 10 = 8 - 24 + 6 + 10 = 0 \Rightarrow x = 2$

    • $f(5) = (5)^3 - 6(5)^2 + 3(5) + 10 = 125 - 150 + 15 + 10 = 0 \Rightarrow x = 5$

  • The $x$-intercepts are $-1$, $2$, and $5$.

• Y-intercept:

  • Set $x = 0$: $y = (0)^3 - 6(0)^2 + 3(0) + 10 = 10$

  • The $y$-intercept is $10$.

Example 2: Testing for Symmetry

Test the curve $y = x^3 - 6x^2 + 3x + 10$ for symmetry.

Solution:

• Symmetry about the y-axis:

  • Substitute $-x$ for $x$: $y = (-x)^3 - 6(-x)^2 + 3(-x) + 10 = -x^3 - 6x^2 - 3x + 10$

  • Since $-x^3 - 6x^2 - 3x + 10 \neq x^3 - 6x^2 + 3x + 10$, there is no symmetry about the $y$-axis.

• Symmetry about the x-axis:

  • Substitute $-y$ for $y$: $-y = x^3 - 6x^2 + 3x + 10 \Rightarrow y = -x^3 + 6x^2 - 3x - 10$

  • Since $-x^3 + 6x^2 - 3x - 10 \neq x^3 - 6x^2 + 3x + 10$, there is no symmetry about the $x$-axis.

• Symmetry about the origin:

  • Substitute $-x$ for $x$ and $-y$ for $y$: $-y = (-x)^3 - 6(-x)^2 + 3(-x) + 10 \Rightarrow y = x^3 + 6x^2 + 3x - 10$

  • Since $x^3 + 6x^2 + 3x - 10 \neq x^3 - 6x^2 + 3x + 10$, there is no symmetry about the origin.

Example 3: Discussing and Tracing a Curve

Discuss and trace the curve $y = x^3 + 3x^2 - 10x - 24$.

Solution:

1. Form of the equation:

   • The equation is already in the form $y = f(x)$, which is a Type 1 curve.

2. Test for symmetry:

   • $x \rightarrow -x$: $y = (-x)^3 + 3(-x)^2 - 10(-x) - 24 = -x^3 + 3x^2 + 10x - 24$. No symmetry with respect to the $y$-axis.

   • $y \rightarrow -y$: $-y = x^3 + 3x^2 - 10x - 24 \Rightarrow y = -x^3 - 3x^2 + 10x + 24$. No symmetry with respect to the $x$-axis.

   • $x \rightarrow -x, y \rightarrow -y$: $-y = (-x)^3 + 3(-x)^2 - 10(-x) - 24 \Rightarrow y = x^3 - 3x^2 + 10x + 24$. No symmetry with respect to the origin.

3. Determine Intersections with the axes:

   • $x$-intercepts: Set $y = 0$:

     • $0 = x^3 + 3x^2 - 10x - 24 = (x - 3)(x + 2)(x + 4)$

     • The curve crosses the $x$-axis at $(3, 0), (-2, 0), (-4, 0)$.

   • $y$-intercepts: Set $x = 0$:

     • $y = (0)^3 + 3(0)^2 - 10(0) - 24 = -24$. The curve crosses the $y$-axis at $(0, -24)$.

4. Determine Asymptotes:

   • Since there is no polynomial denominator, the curve has no asymptotes.

5. Divide the plane into regions:

   • Using the factored form $y = (x - 3)(x + 2)(x + 4)$, analyze the sign:

     • When $x < -4$, e.g., $x = -5$, $y = (-5 - 3)(-5 + 2)(-5 + 4) = -24 < 0$. The curve lies below the $x$-axis.

     • When $-4 < x < -2$, e.g., $x = -3$, $y = (-3 - 3)(-3 + 2)(-3 + 4) = 6 > 0$. The curve lies above the $x$-axis.

     • When $-2 < x < 3$, e.g., $x = 0$, $y = (0 - 3)(0 + 2)(0 + 4) = -24 < 0$. The curve lies below the $x$-axis.

     • When $x > 3$, e.g., $x = 4$, $y = (4 - 3)(4 + 2)(4 + 4) = 48 > 0$. The curve lies above the $x$-axis.

6. Trace the curve:

   • Plotting a few additional points:

     • $(-5, -24), (-4, 0), (-3, 6), (-2, 0), (-1, -12), (0, -24), (1, -30), (2, -24), (3, 0), (4, 48)$

Example 4: Discussing and Tracing a Rational Function

Discuss and trace the curve $x^2y + 2xy - x - 3y + 3 = 0$.

Solution:

1. Form of the equation:

   • Rearrange the equation to solve for $y$:

     • $x^2y + 2xy - 3y = x - 3$

     • $y(x^2 + 2x - 3) = x - 3$

     • $y = \frac{x - 3}{x^2 + 2x - 3} = \frac{x - 3}{(x + 3)(x - 1)}$ (general and factored forms)

   • This is a Type 2 curve.

2. Test for symmetry:

   • $x \rightarrow -x$: $y = \frac{-x - 3}{(-x + 3)(-x - 1)} = \frac{-(x + 3)}{(3 - x)(x + 1)}$ No symmetry with respect to the $y$-axis.

   • $y \rightarrow -y$: $-y = \frac{x - 3}{(x + 3)(x - 1)} \Rightarrow y = -\frac{x - 3}{(x + 3)(x - 1)}$ No symmetry with respect to the $x$-axis.

   • $x \rightarrow -x, y \rightarrow -y$: $-y = \frac{-x - 3}{(-x + 3)(-x - 1)} \Rightarrow y = \frac{x + 3}{(x - 3)(x + 1)}$ No symmetry with respect to the origin.

3. Determine Intersections with the axes:

   • $x$-intercepts: Set $y = 0$:

     • $0 = \frac{x - 3}{(x + 3)(x - 1)} \Rightarrow x = 3$. The curve crosses the $x$-axis at $(3, 0)$.

   • $y$-intercepts: Set $x = 0$:

     • $y = \frac{0 - 3}{(0 + 3)(0 - 1)} = \frac{-3}{-3} = 1$. The curve crosses the $y$-axis at $(0, 1)$.

4. Determine Asymptotes:

   • Vertical asymptotes (where $y \rightarrow \infty$):

     • $(x + 3)(x - 1) = 0 \Rightarrow x = -3, x = 1$

     • The vertical lines $x = -3$ and $x = 1$ are asymptotes.

   • Horizontal asymptote (as $x \rightarrow \infty$):

     • $y = \frac{x - 3}{x^2 + 2x - 3} = \frac{\frac{1}{x} - \frac{3}{x^2}}{1 + \frac{2}{x} - \frac{3}{x^2}}$

     • As $x \rightarrow \infty$, $y = \frac{0 - 0}{1 + 0 - 0} = 0$

     • The horizontal line $y = 0$ (the $x$-axis) is an asymptote.

5. Divide the plane into regions:

   • Using the factored form $y = \frac{x - 3}{(x + 3)(x - 1)}$, consider the vertical asymptotes $x = -3$ and $x = 1$

     • When $x < -3$, e.g., $x = -4$, $y = \frac{-4 - 3}{(-4 + 3)(-4 - 1)} = \frac{-7}{(-1)(-5)} = -\frac{7}{5} < 0$. The curve lies below the $x$-axis.

     • When $-3 < x < 1$, e.g., $x = 0$, $y = \frac{0 - 3}{(0 + 3)(0 - 1)} = \frac{-3}{-3} = 1 > 0$. The curve lies above the $x$-axis.

     • When $1 < x < 3$, e.g., $x = 2$, $y = \frac{2 - 3}{(2 + 3)(2 - 1)} = \frac{-1}{5} = -0.2 < 0$. The curve lies below the $x$-axis.

     • When $x > 3$, e.g., $x = 4$, $y = \frac{4 - 3}{(4 + 3)(4 - 1)} = \frac{1}{21} > 0$. The curve lies above the $x$-axis.

6. Trace the curve:

   • Plotting additional points:

     • $(-5, -0.67), (-4, -1.4), (-2, 1.67), (-1, 1), (0, 1), (2, -0.2), (3, 0), (4, 0.05)$