Day 8 Lines and Graphs Study Notes

Lines and Graphs

• Key terms: y-intercept, x-intercept, origin (0,0).

• Visual cue: a line with positive slope rises from left to right.

• A line is determined by its slope and intercepts; understanding intercepts helps with graphing.

• Slope concept:

  • Slope measures steepness of a line.
  • Positive slope: line goes up as you move left to right.
  • Negative slope: line goes down as you move left to right.
  • Slope formula (rise over run):
  • $m = \frac{\text{rise}}{\text{run}} = \frac{y<em>2 - y</em>1}{x<em>2 - x</em>1}$

• Example: find the slope of the line with points (1,1) and (3,5).

  • $m = \frac{5 - 1}{3 - 1} = \frac{4}{2} = 2$
  • The order of the points does not matter as long as you are consistent: swapping points yields the same slope value.

Slope

• Slope is a measure of steepness; it is the rate of change of y with respect to x.
• Calculation reminder:
• $m = \frac{y<em>2 - y</em>1}{x<em>2 - x</em>1}$
• For the pair (1,1) and (3,5), the slope is m = 2].

Horizontal and Vertical Lines

• Horizontal lines:
  • Equation: y = k where k is a constant.
  • Slope: m = 0.
  • All points on the line have the same y-value (k).
• Vertical lines:
  • Equation: x = k where k is a constant.
  • Slope is undefined (infinite). This corresponds to a vertical rise with zero run.
  • All points on the line have the same x-value (k).

Equations of Lines

• General form of a line: y = mx + b
  • m = slope, b = y-intercept (the value of y where the line crosses the y-axis).
  • Slope-intercept form shows the direct relationship between x and y with the intercept on the y-axis.
• Example: Graph the line given by y = -x + 2.
  • Slope: m = -1$, intercept:$b = 2.

X-Intercept and Y-Intercept

• X-intercept: the point where the line crosses the x-axis; set y = 0 and solve for x.
  • For the line y = -x + 2$, solve$0 = -x + 2$to get$x = 2; x-intercept is (2, 0).
• Example with two points: line through (0, -3) and (4, 0).
  • Slope: m = \frac{0 - (-3)}{4 - 0} = \frac{3}{4}.
  • Equation in slope-intercept form: using (0, -3) gives y = \frac{3}{4}x - 3.
  • X-intercept: set y = 0: 0 = \frac{3}{4}x - 3 \Rightarrow x = 4.
  • Y-intercept: at x = 0, y = -3, so y-intercept is -3.

Point-Slope Form

• If a line has slope m and passes through a point ((x1, y1)), its equation is:
• y - y1 = m(x - x1)
• Example: Find the equation of the line through ((-1, 1)) and ((3, 7)).
  • Slope: m = \frac{7 - 1}{3 - (-1)} = \frac{6}{4} = \frac{3}{2}.
  • Using point ((-1, 1)):
  • y - 1 = \frac{3}{2}(x - (-1)) = \frac{3}{2}(x + 1).
  • To slope-intercept form, simplify:
  • y = \frac{3}{2}x + \frac{5}{2}.

Converting to Slope-Intercept Form

• From point-slope: y - y1 = m(x - x1), solve for y:
• y = mx + (y1 - m x1).
• Example recap: with (m = \frac{3}{2}), (x1 = -1), (y1 = 1):
• Intercept b = y1 - m x1 = 1 - \frac{3}{2}(-1) = 1 + \frac{3}{2} = \frac{5}{2}.
• Therefore, y = \frac{3}{2}x + \frac{5}{2}.

Standard Form

• Standard form: A x + B y = C where A, B, C are integers (often with gcd(A,B,C) = 1).
• To find slope from standard form, solve for y to get slope-intercept form:
• A x + B y = C \Rightarrow B y = -A x + C \Rightarrow y = -\frac{A}{B}x + \frac{C}{B}.
• The slope is m = -\frac{A}{B}.
• Example: 2x + 3y = 1.
  • Solve for y: 3y = -2x + 1 \Rightarrow y = -\frac{2}{3}x + \frac{1}{3}.
  • Slope: m = -\frac{2}{3}.

Applied Example: Linear Depreciation Model

• Problem: Mark bought a car for $20,000 and it will have a trade-in value of $6,000 in 10 years. Assuming a constant rate of depreciation (linear), find a linear model describing the value after t years.
• Given:
  • When t = 0$,$V = 20{,}000.
  • When t = 10$,$V = 6{,}000.
• Slope (rate of depreciation):
  • m = \frac{6{,}000 - 20{,}000}{10 - 0} = \frac{-14{,}000}{10} = -1{,}400.
• Intercept (value at t = 0): b = 20{,}000.
• Linear model (value as a function of time):
  • V(t) = -1{,}400\, t + 20{,}000.
• Verification:
  • V(0) = -1{,}400\cdot 0 + 20{,}000 = 20{,}000.
  • V(10) = -1{,}400\cdot 10 + 20{,}000 = -14{,}000 + 20{,}000 = 6{,}000.

Key Takeaways and Connections

• The slope m is the cornerstone of linear relationships; it dictates direction and steepness of the graph.
• Intercepts, both x- and y-, anchor the line on the axes and aid in quick graphing.
• Multiple forms exist for representing lines:
  • Slope-intercept: y = mx + b, emphasizes slope and y-intercept.
  • Point-slope: y - y1 = m(x - x1), useful when a point on the line is known.
  • Standard form: A x + B y = C, useful for certain algebraic manipulations and systems.
• Converting between forms involves algebraic rearrangement, with the slope reminding that m = -\frac{A}{B} when in standard form.
• Real-world relevance: linear models describe constant-rate changes such as depreciation, simple interest, and basic cost-revenue relationships.
• Ethical/practical note: choosing the right model and understanding its limits is essential; linear models assume constant rate of change which may not hold for long horizons or nonlinear phenomena.

Quick Practice Recap

• Slope between two points: m = \frac{y2 - y1}{x2 - x1}.
• Horizontal line: y = k\Rightarrow m = 0.
• Vertical line: x = k\Rightarrow m\text{ undefined}.
• X-intercept of line y = mx + b$: set y = 0, solve for x:$0 = mx + b \Rightarrow x = -\frac{b}{m} (when m ≠ 0).
• Standard form to slope-intercept: y = -\frac{A}{B}x + \frac{C}{B}$for$A x + B y = C.
• Depreciation model example: V(t) = -1400 t + 20000.$$