Notes on Linear Equations in Two Variables, Slopes, and Functions

Graphing Linear Equations in Two Variables (Sections 2.1–2.2)

• A linear equation in two variables has infinitely many solutions.

  • Each solution is a coordinate pair (x, y) in the Cartesian plane, not a single number.

  • The set of all solutions forms a line when graphed on the plane.

• Graphing the solution set:

  • To graph a line, you only need two points that lie on the line.

  • A convenient pair of points are the intercepts:

  • x-intercept: the point where the line crosses the x-axis, obtained by setting y = 0 and solving for x; the point is
    $(x, 0)$.

  • y-intercept: the point where the line crosses the y-axis, obtained by setting x = 0 and solving for y; the point is
    $(0, y)$.

  • Note on notation: some textbooks refer to the intercepts by the numbers themselves (e.g., the x-intercept might be denoted by the value a in a point (a, 0) or simply a).

• Example: graphing the line from $3x+4y=12$

  • x-intercept: set $y=0$ → $3x=12 \,\Rightarrow\ x=4$ ⇒ intercept point $(4,0)$.

  • y-intercept: set $x=0$ → $4y=12 \,\Rightarrow\ y=3$ ⇒ intercept point $(0,3)$.

  • Plot these two points and draw the line through them.

  • Therefore, the graph is the line corresponding to the equation $3x+4y=12$.

• Another example (sketch context): for the line given by $-2x+5=0$ (interpreting in the intercept sense with y set to 0):

  • x-intercept: set $y=0$ → $-2x+5=0\Rightarrow x=2.5$ ⇒ intercept $(2.5,0)$.

  • y-intercept: set $x=0$ → $y=5$ ⇒ intercept $(0,5)$.

  • Plot and connect to obtain the line for $-2x+5=0$.

• A special case: horizontal and vertical lines

  • If the equation is of the form $y=c$, the graph is a horizontal line through the point $(0,c)$ (the y-intercept is $(0,c)$).

  • If the equation is of the form $x=c$, the graph is a vertical line through the point $(c,0)$ (the x-intercept is $(c,0)$).

  • Example: $y=3$ graphs as a horizontal line; all points on the line have y = 3.

  • Example: $x=-1$ graphs as a vertical line; all points on the line have x = -1.

• Finding an equation for a line from slope and a point

  • If you know two points $(x<em>1,y</em>1)$ and $(x<em>2,y</em>2)$ with $x<em>1 \neq x</em>2$, the slope is
    $m = \frac{y<em>2 - y</em>1}{x<em>2 - x</em>1}.$

  • If you know a point $(x<em>1,y</em>1)$ and the slope $m$, you can use the point-slope form to write the equation:
    $y - y<em>1 = m\,(x - x</em>1).$

  • Converting to slope-intercept form gives $y = m x + b$ with $b = y<em>1 - m x</em>1$.

• Quick illustration of point-slope to slope-intercept with an example

  • Line with slope $m=2$ through $(3,5)$:

  $y - 5 = 2\,(x - 3) \quad\Rightarrow\quad y = 2x - 1.$

  • Line with slope $m=6$ through $(-2,5)$:

  $y - 5 = 6\,(x + 2) \Rightarrow y = 6x + 17.$

• The slope-intercept form and using the intercept

  • If the line has y-intercept $b$ and slope $m$, the equation is
    $y = m x + b.$

  • Derivation (brief): from the slope definition, $m = \dfrac{y - b}{x - 0} = \dfrac{y - b}{x}$, hence $y = m x + b$.

• Sketching tips using slope (rise over run)

  • From the y-intercept $(0,b)$, use the slope $m = \dfrac{\text{rise}}{\text{run}}$; for example, a slope of $3$ can be drawn as rise 3, run 1.

  • Move from the intercept point accordingly to place another point on the line, then draw through.

• Parallel and perpendicular lines

  • Non-vertical lines: $l<em>1$ with slope $m</em>1$ and $l<em>2$ with slope $m</em>2$ are parallel if $m<em>1 = m</em>2$.

  • Perpendicular: $m<em>1 m</em>2 = -1$ (provided neither line is vertical).

  • Example reasoning: if two lines have the same slope, they never intersect; if their slopes multiply to $-1$, they intersect at a right angle.

Determining parallel or perpendicular relationships (Example 8)

• Approach: solve each line for $y$ in terms of $x$ to identify the slope, since the slope is the coefficient of $x$ in the form $y = m x + b.$

• Part a

  • L1: $2x + 3y = 5 \Rightarrow 3y = -2x + 5 \Rightarrow y = -\frac{2}{3}x + \frac{5}{3}$ → $m_1 = -\frac{2}{3}$.

  • L2: $4x + 6y = -5 \Rightarrow 6y = -4x - 5 \Rightarrow y = -\frac{4}{6}x - \frac{5}{6} = -\frac{2}{3}x - \frac{5}{6}$ → $m_2 = -\frac{2}{3}$.

  • Since $m<em>1 = m</em>2$, the lines are parallel.

• Part b

  • L1: $3x = y + 7 \Rightarrow y = 3x - 7$ → $m_1 = 3$.

  • L2: $x + 3y = 4 \Rightarrow 3y = -x + 4 \Rightarrow y = -\frac{1}{3}x + \frac{4}{3}$ → $m_2 = -\frac{1}{3}$.

  • Product: $m<em>1 m</em>2 = 3 \cdot (-\frac{1}{3}) = -1$ → The lines are perpendicular.

• Part c: left as exercise (check if $m<em>1 = m</em>2$ or if $m<em>1 m</em>2 = -1$; otherwise neither).

Functions and the Graph of a Function (Section 2.3)

• A function is a relation between two quantities, with

  • inputs (domain): the set of inputs you plug into the function,

  • outputs (range): the set of possible outputs produced by the function,

  • a rule that assigns to each input exactly one output.

• Notation and terminology

  • Inputs are called the independent variable (commonly $x$).

  • Outputs are called the dependent variable (commonly $y$).

  • A function may be named $f$, $g$, $h$, etc., and written as $y = f(x).$

• Examples of functions (non-algebraic domain intuition)

  • Domain: Virginia Tech students; Range: unique passport IDs. Each student maps to exactly one passport ID.

  • Domain: graduating seniors; Range: GPAs. The GPA is a function of the student (subject to the graduation GPA rule, typically $0\le \text{GPA} \le 4.0$, with graduation requiring GPA at least $2.0$ in the example).

• Algebraic function (formulas)

  • Typical setup: domain and range are the real numbers; a function is described by a rule such as
    $y = f(x).$

  • Example: $y = 2x + 1$ describes a function from real numbers to real numbers where input $x$ is the independent variable and output is $y$ (the dependent variable).

• Example 4: a concrete formula

  • Given $f(x) = x^2 - 2x + 3.$

  • Part a: compute $f(1)$:
    $f(1) = 1^2 - 2(1) + 3 = 1 - 2 + 3 = 2.$

  • Part b: compute $f(5)$:
    $f(5) = 5^2 - 2(5) + 3 = 25 - 10 + 3 = 18.$

  • Part c: general input $a$:
    $f(a) = a^2 - 2a + 3.$

  • Part d: input $a + h$:
    $f(a + h) = (a + h)^2 - 2(a + h) + 3 = a^2 + 2ah + h^2 - 2a - 2h + 3.$

  • Takeaway: a single formula can describe the same function; the expression can be expanded for different inputs.

• Piecewise-defined functions

  • A function can be defined by different formulas on different parts of the domain.

  • Example (Example 5 in the transcript):
    f(x) = egin{cases} frac{1}{2}x^2 + 1, & x < 2,\ x^2 + 5, & x \ge 2. \ frac{?}{?} & ext{(typo in transcript)} \ \ ext{end cases} \
    \

  • Using this piecewise definition, evaluate:

  • f(-2)$(since$-2 < 2$, use the top formula):<br>$f(-2) = \tfrac{1}{2}(-2)^2 + 1 = \tfrac{1}{2} \cdot 4 + 1 = 2 + 1 = 3.$</p></li><li><p>$f(0)$(still$0 < 2$):<br>$f(0) = \tfrac{1}{2}(0)^2 + 1 = 0 + 1 = 1.$</p></li><li><p>$f(2)$(now$2 \ge 2$, use bottom formula):<br>$f(2) = 2^2 + 5 = 4 + 5 = 9.$</p></li><li><p>$f(4)$(bottom branch):<br>$f(4) = 4^2 + 5 = 16 + 5 = 21.$$

• Final note

  • Piecewise functions allow modeling of different rules on different parts of the domain, which the transcript demonstrates with a two-branch example.

• This concludes the sections 2.1, 2.2, and an overview of 2.3; next time we’ll continue with more on functions and their graphs.