Task 3.2: Algebraic Verification (Equation 2) (5 Marks) Verification through algebra: confirm the y-intercept below to show your working.

The equations $y = \frac{1}{2}x + 2$, $y = 2x - 4$, and $y = -2x + 2$ are all in the form $y = mx + c$. This is the standard form for a linear equation.

When graphed, equations of the form $y = mx + c$ always produce straight lines.

• Answer: The equations are linear because their graphs are straight lines.

Phase 3: Investigating Intercepts (12 Marks)

Task 3.1: Visual Identification (6 Marks)

• Equation 1: $y = \frac{1}{2}x + 2$

  • y-intercept ($x=0$): $y = \frac{1}{2}(0) + 2 = 2$. Ordered pair: $(0; 2)$.
  • x-intercept ($y=0$): $0 = \frac{1}{2}x + 2 \implies -2 = \frac{1}{2}x \implies x = -4$. Ordered pair: $(-4; 0)$.

• Equation 2: $y = 2x - 4$

  • y-intercept ($x=0$): $y = 2(0) - 4 = -4$. Ordered pair: $(0; -4)$.
  • x-intercept ($y=0$): $0 = 2x - 4 \implies 4 = 2x \implies x = 2$. Ordered pair: $(2; 0)$.

• Equation 3: $y = -2x + 2$

  • y-intercept ($x=0$): $y = -2(0) + 2 = 2$. Ordered pair: $(0; 2)$.
  • x-intercept ($y=0$): $0 = -2x + 2 \implies 2x = 2 \implies x = 1$. Ordered pair: $(1; 0)$.

Task 3.2: Algebraic Verification (Equation 2) (4 Marks)

Workspace for Equation 2 ($y = 2x - 4$):

The y-intercept is $(0; -4)$.

The x-intercept is $(2; 0)$.

Task 3.3: Synthesis of Intercepts (2 Marks)

The y-intercept is the point where the graph crosses the y-axis. At this point, the x-coordinate is always zero.

For a linear equation in the form $y = mx + c$, the y-intercept is the point $(0, c)$, where $c$ is the constant term.

• Answer: The y-intercept is the point where the graph intersects the y-axis. Its x-coordinate is always 0, and its y-coordinate is the value of the function when $x=0$. In the equation $y = mx + c$, the y-intercept is $(0, c)$.