Notes on Cartesian Planes, Graphs, and Linear Models

Cartesian Planes (Page 1)

  • The regular Cartesian plane is the coordinate grid used in this material.
  • Named for Rene Descartes (French mathematician, philosopher, scientist).
  • Attributes of the Cartesian plane:
    • Two real number lines (typically the x- and y-axes); can also be used to compare any two quantities.
    • The point where the axes intersect is the origin.
    • The number lines are perpendicular.
    • The axes divide the plane into four quadrants.
  • Which is the correct form for an ordered pair?
    • Correct: (x,y)(x, y)
    • Incorrect: (y,x)(y, x)
  • Plot the points: A$(5, -3)$, B$(-3, 8)$, C$(0, -3i)$
    • Note: C has an imaginary component ($-3i$); in the real Cartesian plane, coordinates are real numbers, so C is not a standard real point.

Distance Formula and Pythagoras (Page 2)

  • Distance formula (Pythagorean theorem in disguise):
    • d=(x<em>2−x</em>1)2+(y<em>2−y</em>1)2d = \sqrt{(x<em>2 - x</em>1)^2 + (y<em>2 - y</em>1)^2}
  • Pythagoras: c2=a2+b2c^2 = a^2 + b^2
  • Example with Minesweepers (Battleship analogy):
    • Minesweeper 1 at $(4, 7)$ and Minesweeper 2 at $(-1, -3)$; each grid space is a 1.5 cm square.
    • Distance in grid units: d=(x<em>2−x</em>1)2+(y<em>2−y</em>1)2=(−1−4)2+(−3−7)2=(−5)2+(−10)2=25+100=125=55.d = \sqrt{(x<em>2 - x</em>1)^2 + (y<em>2 - y</em>1)^2} = \sqrt{(-1 - 4)^2 + (-3 - 7)^2} = \sqrt{(-5)^2 + (-10)^2} = \sqrt{25 + 100} = \sqrt{125} = 5\sqrt{5}.\
    • Convert to centimeters: dcm=d⋅1.5=55⋅1.5 cm≈16.77 cm.d_{\text{cm}} = d \cdot 1.5 = 5\sqrt{5} \cdot 1.5 \text{ cm} \approx 16.77\text{ cm}.
  • Midpoint formula (averaging):
    • Midpoint=(x<em>1+x</em>22,y<em>1+y</em>22)\text{Midpoint} = \left(\dfrac{x<em>1 + x</em>2}{2}, \dfrac{y<em>1 + y</em>2}{2}\right)
    • Two uses:
      1) Find the midpoint of a segment connecting $(x1,y1)$ and $(x2,y2)$.
      2) Use the midpoint as a key point in other constructions or checks (e.g., perpendicular bisectors, symmetry tests).

Campus Location Problem and Endpoint From Midpoint (Page 3)

  • Equal traveling distance problem (Smithstown and Baskerville relative to intersection):
    • Smithstown is 4 miles south and 7 miles east of the intersection: S = $(7, -4)$.
    • Baskerville is 8 miles north and 1 mile west of the intersection: B = $(-1, 8)$.
    • To be equidistant from S and B, the campus must lie on the perpendicular bisector of the segment SB.
    • Midpoint of SB: M=(7+(−1)2,−4+82)=(3, 2).M = \left(\dfrac{7 + (-1)}{2}, \dfrac{-4 + 8}{2}\right) = (3,\,2).
    • Slope of SB: m{SB} = \dfrac{8 - (-4)}{-1 - 7} = \dfrac{12}{-8} = -\dfrac{3}{2}.$n- Perpendicular slope: m{\perp} = -\dfrac{1}{m_{SB}} = \dfrac{2}{3}.
    • Perpendicular bisector line through $M$: y - 2 = \dfrac{2}{3}(x - 3).
    • Check whether the intersection point (the origin) lies on this line:
    • If $(0,0)$, then $0 - 2 = (2/3)(0 - 3) = -2$, which holds; so the origin is on the perpendicular bisector.
    • Therefore, the campus should be built at the intersection point (the origin) to ensure equal traveling distance to both towns.
  • Endpoints given midpoint problem:
    • Endpoint is at $(-2, 8)$ and the midpoint is at $(1, -3)$. Let the other endpoint be $(x2, y2)$.
    • Use the midpoint formula:
    • \dfrac{-2 + x2}{2} = 1 \Rightarrow x2 = 4,</li><li></li> <li>\dfrac{8 + y2}{2} = -3 \Rightarrow y2 = -14.
    • Therefore, the other endpoint is $(4, -14)$.
  • Graphs of Equations: Determining solution points
    • Given the function (equation not fully shown): Is $(2,13)$ a solution? For $y = 10x - 7$:
    • Evaluate: $y = 10(2) - 7 = 20 - 7 = 13$; Yes, $(2,13)$ is a solution.
    • Is $(-1, -3)$ a solution?
    • Evaluate: $y$ predicted by the equation at $x = -1$ is $10(-1) - 7 = -17$; since $y = -3$ in the point, it is not a solution.
    • How you know: A point is on the graph if and only if its coordinates satisfy the equation.

Sketching Graphs of Functions (Page 4)

  • Goal: create a sketch without a t-chart; rely on the graphing calculator.
  • Important points often require precise plotting or labeling: vertex of a parabola, maximum, minimum, center of a circle, x-intercept, y-intercept, and asymptotes.
  • You may need to search for how to find these features on a TI-84+ CE (YouTube resources suggested in the material).
  • Graph the following functions:
    • y = x^{2}</li><li></li> <li>y = 1 - (x - 5)^{2}</li><li></li> <li>y = x^{3} - 4x</li><li></li> <li>y = |2x|</li><li></li> <li> (x + 1)^{2} + (y - 2)^{2} = 20
  • Important note / correction:
    • The statement "$y = |2x|$ → this is a circle; both variables (x and y) are positive and squared" is incorrect.
    • In fact, $y = |2x|$ is a V-shaped graph with vertex at the origin. The circle is described by equations such as $(x + 1)^{2} + (y - 2)^{2} = 20$.
  • What must you change to graph on the calculator?
    • You need to switch to graphing mode and enter the functions in the suitable input locations (e.g., Y1, Y2, etc.) so that the TI-84+ CE can plot them.
    • Ensure you are using the correct form of each equation and appropriate window settings if needed.

Symmetry in Graphs and Slope Concepts (Page 5)

  • Symmetry rules:
    1) If both $(x, y)$ and $(x, -y)$ lie on the graph, the graph has symmetry with respect to the x-axis. Visually, the graph can fold over itself across the x-axis.
    2) If both $(x, y)$ and $(-x, y)$ lie on the graph, the graph has symmetry with respect to the y-axis. Visually, the graph can fold over itself across the y-axis.
    3) If both $(x, y)$ and $(-x, -y)$ lie on the graph, the graph has symmetry with respect to the origin (rotation by 180 degrees).
  • There are other axes of symmetry as well: lines such as $y = a$, $x = b$, $y = x$, and $y = -x$ are examples; in fact there are infinitely many lines of symmetry for appropriate curves.
  • Linear equations in two variables: Using/finding slope
    • What is slope? A measure of the steepness of a line; can be interpreted as "rise over run".
    • Four types of slope (conceptual):
      1) Positive slope (increases as x increases)
      2) Negative slope (decreases as x increases)
      3) Zero slope (horizontal line, dy/dx = 0)
      4) Undefined slope (vertical line, undefined because division by zero in run)

Point-Slope Form and Parallel/Perpendicular Lines (Page 6)

  • Point-slope form:
    • y - y{1} = m(x - x{1})
    • Example: Write, in point-intercept form, the equation with a slope of $4$ and $(x{1}, y{1}) = (5, -2)$:
    • Using point-slope form: y - (-2) = 4(x - 5) \Rightarrow y + 2 = 4x - 20.</li><li>Slope−interceptformthenis:</li> <li>Slope-intercept form then is:y = 4x - 22.
  • Parallel and perpendicular lines:
    • Parallel lines have the same slope as each other (equal slopes).
    • Perpendicular lines have slopes that are negative reciprocals of each other (product of slopes is $-1$).
  • Find the slope-intercept form for the equation of a line that is parallel AND perpendicular to the given line with a given point:
    • Given point $P(3,5)$ and line $2x - 3y = 5$:
    • Rewrite the given line in slope-intercept form to read the slope: 2x - 3y = 5 \Rightarrow -3y = -2x + 5 \Rightarrow y = \frac{2}{3}x - \frac{5}{3}.</li><li>Slopeofthegivenline:</li> <li>Slope of the given line:m_{g} = \frac{2}{3}.
    • Parallel line through $P$: slope $m_{p} = \frac{2}{3}$, equation in slope-intercept form:
      • Start with point-slope: y - 5 = \frac{2}{3}(x - 3).</li><li>Convert:</li> <li>Convert:y = \frac{2}{3}x + 3. (Parallel to the given line)
    • Perpendicular line through $P$: slope $m{\perp} = -\frac{1}{m{g}} = -\frac{3}{2}$, equation in slope-intercept form:
      • Start with point-slope: y - 5 = -\frac{3}{2}(x - 3).</li><li>Convert:</li> <li>Convert:y = -\frac{3}{2}x + \frac{19}{2}. (Perpendicular to the given line)

Applications: Linear Models and Depreciation/Revenue (Page 7)

  • Linear depreciation model (t = years since purchase):
    • Given: initial value $V(0) = 5000$, value after 10 years $V(10) = 400$.
    • Slope: m = \dfrac{V(10) - V(0)}{10 - 0} = \dfrac{400 - 5000}{10} = -460.</li><li>Model:</li> <li>Model:V(t) = 5000 - 460t.</li><li>Valueafter7years:</li> <li>Value after 7 years:V(7) = 5000 - 460 \cdot 7 = 5000 - 3220 = 1780.
  • Revenue projection using a linear model:
    • Given: year 3 revenue $R(3) = 124{,}034$, year 5 revenue $R(5) = 150{,}348$.
    • Slope: m = \dfrac{150{,}348 - 124{,}034}{5 - 3} = \dfrac{26{,}314}{2} = 13{,}157.
    • Intercept (use $R(x) = mx + b$ with $x=3$, $R=124{,}034$): 124{,}034 = 13{,}157 \times 3 + b \Rightarrow b = 84{,}563.</li><li>Projectedrevenueforyear11:</li> <li>Projected revenue for year 11:R(11) = 13{,}157 \times 11 + 84{,}563 = 144{,}727 + 84{,}563 = 229{,}290.$$