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)(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)(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=123x+4y=12

    • x-intercept: set $y=0$ → 3x=12 x=43x=12 \,\Rightarrow\ x=4 ⇒ intercept point (4,0)(4,0).

    • y-intercept: set $x=0$ → 4y=12 y=34y=12 \,\Rightarrow\ y=3 ⇒ intercept point (0,3)(0,3).

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

    • Therefore, the graph is the line corresponding to the equation 3x+4y=123x+4y=12.

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

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

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

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

  • A special case: horizontal and vertical lines

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

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

    • Example: x=1x=-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 $(x1,y1)$ and $(x2,y2)$ with $x1 \neq x2$, the slope is
      m=y<em>2y</em>1x<em>2x</em>1.m = \frac{y<em>2 - y</em>1}{x<em>2 - x</em>1}.

    • If you know a point $(x1,y1)$ and the slope $m$, you can use the point-slope form to write the equation:
      yy<em>1=m(xx</em>1).y - y<em>1 = m\,(x - x</em>1).

    • Converting to slope-intercept form gives y=mx+by = m x + b with $b = y1 - m x1$.

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

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

    y5=2(x3)y=2x1.y - 5 = 2\,(x - 3) \quad\Rightarrow\quad y = 2x - 1.

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

    y5=6(x+2)y=6x+17.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=mx+b.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: $l1$ with slope $m1$ and $l2$ with slope $m2$ are parallel if $m1 = m2$.

    • Perpendicular: $m1 m2 = -1$ (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=mx+b.y = m x + b.

  • Part a

    • L1: 2x+3y=53y=2x+5y=23x+532x + 3y = 5 \Rightarrow 3y = -2x + 5 \Rightarrow y = -\frac{2}{3}x + \frac{5}{3} → $m_1 = -\frac{2}{3}$.

    • L2: 4x+6y=56y=4x5y=46x56=23x564x + 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 $m1 = m2$, the lines are parallel.

  • Part b

    • L1: 3x=y+7y=3x73x = y + 7 \Rightarrow y = 3x - 7 → $m_1 = 3$.

    • L2: x+3y=43y=x+4y=13x+43x + 3y = 4 \Rightarrow 3y = -x + 4 \Rightarrow y = -\frac{1}{3}x + \frac{4}{3} → $m_2 = -\frac{1}{3}$.

    • Product: $m1 m2 = 3 \cdot (-\frac{1}{3}) = -1$ → The lines are perpendicular.

  • Part c: left as exercise (check if $m1 = m2$ or if $m1 m2 = -1$; 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).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).y = f(x).

    • Example: y=2x+1y = 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)=x22x+3.f(x) = x^2 - 2x + 3.

    • Part a: compute $f(1)$:
      f(1)=122(1)+3=12+3=2.f(1) = 1^2 - 2(1) + 3 = 1 - 2 + 3 = 2.

    • Part b: compute $f(5)$:
      f(5)=522(5)+3=2510+3=18.f(5) = 5^2 - 2(5) + 3 = 25 - 10 + 3 = 18.

    • Part c: general input $a$:
      f(a)=a22a+3.f(a) = a^2 - 2a + 3.

    • Part d: input $a + h$:
      f(a+h)=(a+h)22(a+h)+3=a2+2ah+h22a2h+3.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):
      f(-2) = \tfrac{1}{2}(-2)^2 + 1 = \tfrac{1}{2} \cdot 4 + 1 = 2 + 1 = 3.

    • $f(0)$ (still $0 < 2$):
      f(0) = \tfrac{1}{2}(0)^2 + 1 = 0 + 1 = 1.

    • $f(2)$ (now $2 \ge 2$, use bottom formula):
      f(2) = 2^2 + 5 = 4 + 5 = 9.

    • $f(4)$ (bottom branch):
      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.