Graphing Linear Equations: Point-Slope and Slope-Intercept Analysis

Point-Slope Form and Problem Specifications

• Given Parameters:

  • Slope ($m$): $m = 4$
  • Given Point ($(x_1, y_1)$): $(-2, -5)$
  • Independent variable coordinate ($x_1$): $-2$
  • Dependent variable coordinate ($y_1$): $-5$

• Point-Slope Form Definition:

  • The point-slope form of a linear equation is derived directly from the definition of slope: $m = \frac{y - y_1}{x - x_1}$
  • Multiplying both sides by the denominator yields the standard point-slope formula: $y - y_1 = m(x - x_1)$

• Substitution and Algebraic Formulation:

  • Substitute the given coordinates and slope into the formula: $y - (-5) = 4(x - (-2))$
  • Simplifying double negatives yields the point-slope equation: $y + 5 = 4(x + 2)$

Alternative Algebraic Forms

• Slope-Intercept Form ($y = mx + b$):

  • Distribute the slope across the parentheses: $y + 5 = 4x + 8$
  • Subtract $5$ from both sides to isolate $y$: $y = 4x + 3$
  • Slope ($m$): $4$
  • $y$-intercept ($b$): $3$, corresponding to the coordinate point $(0, 3)$

• Standard Form ($Ax + By = C$):

  • Rearrange $y = 4x + 3$ by subtracting $4x$ from both sides: $-4x + y = 3$
  • Multiply by $-1$ to ensure $A \ge 0$: $4x - y = -3$
  • Coefficients: $A = 4$, $B = -1$, $C = -3$

• Intercept Determinations:

  • $y$-intercept: Set $x = 0$ in $y = 4(0) + 3$, yielding $y = 3$; coordinate is $(0, 3)$
  • $x$-intercept: Set $y = 0$ in $0 = 4x + 3$, yielding $4x = -3 \implies x = -\frac{3}{4} = -0.75$; coordinate is $\\left(-\frac{3}{4}, 0\\right)$

Graphing Procedure and Coordinate Grid Analysis

• Grid Dimensions and Domain/Range:

  • The coordinate plane grid spans from $-5$ to $5$ on the horizontal $x$-axis: domain $[-5, 5]$
  • The coordinate plane grid spans from $-5$ to $5$ on the vertical $y$-axis: range $[-5, 5]$
  • Grid increment spacing is $1$ unit per subdivision along both axes.

• Step-by-Step Graph Construction:

  • Step 1: Plot the Initial Point: Locate the given coordinate $(-2, -5)$ in the third quadrant (two units left on the $x$-axis, five units down on the $y$-axis) and plot a solid point.
  • Step 2: Interpret the Slope: Interpret the slope $m = 4$ as a ratio of vertical change (rise) to horizontal change (run): $m = \frac{\Delta y}{\Delta x} = \frac{4}{1}$
  • Step 3: Plot Subsequent Points via Rise/Run:
    • From $(-2, -5)$, move up $4$ units and right $1$ unit to locate $(-2 + 1, -5 + 4) = (-1, -1)$.
    • From $(-1, -1)$, move up $4$ units and right $1$ unit to locate the $y$-intercept: $(-1 + 1, -1 + 4) = (0, 3)$.
    • The next integer point along this trajectory would be $(0 + 1, 3 + 4) = (1, 7)$, which falls outside the vertical boundary of the $[-5, 5]$ grid.
  • Step 4: Draw the Line: Align a straightedge through the plotted collinear points $(-2, -5)$, $(-1, -1)$, and $(0, 3)$. Draw a continuous straight line extending to the edges of the grid, placing arrowheads at both ends to indicate infinite extension in both directions.

• Grid Boundary Intersection Points:

  • Lower boundary intersection: At $y = -5$, $x = -2$, which coincides exactly with the given anchor point $(-2, -5)$.
  • Upper boundary intersection: Set $y = 5$ in $5 = 4x + 3 \implies 4x = 2 \implies x = 0.5$. The line exits the top of the visible grid at $(0.5, 5)$.
  • The visible line segment on this grid exists strictly over the domain $[-2, 0.5]$ and range $[-5, 5]$.

Geometric Properties of the Line

• Rate of Change:

  • The line exhibits a strictly positive slope ($m > 0$), indicating that the function is strictly monotonically increasing across its entire domain.
  • For every unit increase in $x$, $y$ increases by $4$ units.

• Direction and Angle of Inclination:

  • The angle of inclination $\theta$ with respect to the positive $x$-axis is given by: $\tan(\theta) = m = 4$
  • $\theta = \arctan(4) \approx 75.96^\circ$ (or approximately $1.326\,\text{rad}$), demonstrating a steep upward trajectory.