Fundamental Concepts in College Algebra: Distance, Midpoint, Graphing, and Linear Equations

Exercises for Fundamentals Chapter Section 1

Distance Between Two Points

The distance formula is used to find the length between two points in the Cartesian plane. This is a foundational concept in analytic geometry and College Algebra.

• Definition: The distance between points $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$ is given by:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

• Application: Use the formula to compute the distance for given pairs of points.

• Example: For $P_1 = (-2, -3)$ and $P_2 = (4, 7)$, substitute into the formula:

$d = \sqrt{(4 - (-2))^2 + (7 - (-3))^2} = \sqrt{6^2 + 10^2} = \sqrt{36 + 100} = \sqrt{136}$

Midpoint of a Line Segment

The midpoint formula finds the point exactly halfway between two given points in the plane.

• Definition: The midpoint $M$ between $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$ is:

$M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

• Application: Use the formula to find the midpoint for given pairs of points.

• Example: For $P_1 = (-4, -3)$ and $P_2 = (2, 3)$:

$M = \left( \frac{-4 + 2}{2}, \frac{-3 + 3}{2} \right) = (-1, 0)$

Poverty Thresholds and Mathematical Modeling

Mathematical modeling uses algebraic equations to represent real-world scenarios, such as poverty thresholds determined by the U.S. Census Bureau.

• Definition: A poverty threshold is the minimum annual income for a family to be considered poor.

• Application: Use given data to create a linear model relating family size to poverty threshold.

• Example: If the poverty threshold for a family of four is $22,831 and for a family of five is $26,170, a linear equation can be constructed to estimate the threshold for other family sizes.

Additional info: Students may be asked to use the two-point form of a line to model the relationship.

Exercises for Fundamentals Chapter Section 2

Intercepts and Graphing Equations

Intercepts are points where a graph crosses the axes. Finding intercepts is essential for graphing equations.

• x-intercept: Set $y = 0$ and solve for $x$.

• y-intercept: Set $x = 0$ and solve for $y$.

• Example: For $y = 2x - 3$, the y-intercept is $-3$ and the x-intercept is $x = \frac{3}{2}$.

Testing for Symmetry

Symmetry helps determine the shape and properties of a graph.

• Symmetry about the y-axis: Replace $x$ with $-x$; if the equation is unchanged, the graph is symmetric about the y-axis.

• Symmetry about the x-axis: Replace $y$ with $-y$; if unchanged, symmetric about the x-axis.

• Symmetry about the origin: Replace $x$ with $-x$ and $y$ with $-y$; if unchanged, symmetric about the origin.

• Example: For $y = x^2$, replacing $x$ with $-x$ gives $y = (-x)^2 = x^2$, so the graph is symmetric about the y-axis.

Exercises for Fundamentals Chapter Section 3

Equations of Lines

Linear equations describe straight lines in the plane. The equation can be written in several forms.

• Slope-intercept form: $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

• Point-slope form: $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line.

• Example: Find the equation of a line through $(1, -2)$ with slope $3$:

$y + 2 = 3(x - 1)$

Cost Functions and Modeling

Cost functions model the expenses of running a business. Fixed costs do not change with production level, while variable costs do.

• Fixed cost: Cost that remains constant regardless of output.

• Variable cost: Cost that changes with the level of production.

• Total cost: $\text{Total Cost} = \text{Fixed Cost} + \text{Variable Cost}$

• Example: If fixed cost is $800$and variable cost per unit is $5$, for $x$ units:

$\text{Total Cost} = 800 + 5x$

Equations of Circles

The equation of a circle in the plane is based on its center and radius.

• Standard form: $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.

• Example: For center $(1, 2)$ and radius $3$:

$(x - 1)^2 + (y - 2)^2 = 9$

Table: Summary of Key Formulas

Concept	Formula	Example
Distance between points	$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$	$P_1 = (1,2), P_2 = (4,6): d = \sqrt{(4-1)^2 + (6-2)^2} = 5$
Midpoint	$M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$	$P_1 = (1,2), P_2 = (4,6): M = (2.5, 4)$
Circle equation	$(x - h)^2 + (y - k)^2 = r^2$	Center $(1,2)$, radius $3$:$(x-1)^2 + (y-2)^2 = 9$
Line equation (point-slope)	$y - y_1 = m(x - x_1)$	Through $(1,2)$, slope $3$:$y - 2 = 3(x - 1)$