Revision (Grade 12 NSC Matric Mathematics): Revision Notes

Worked Example: Linear Function

Question: Draw a graph of $2y + x - 8 = 0$ and find the significant characteristics of this linear function.

Solution:

Step 1: Write the equation in standard form $y = mx + c$

Starting with: $2y + x - 8 = 0$

Rearrange: $2y = -x + 8$

Divide by 2: $y = -\frac{1}{2}x + 4$

Therefore: $m = -\frac{1}{2}$ and $c = 4$

Step 2: Draw the straight line graph

Using the gradient-intercept method:

• y-intercept: $(0, 4)$
• Gradient: $m = -\frac{1}{2}$

Alternative method - find both intercepts:

• For y-intercept, let $x = 0$: $y = -\frac{1}{2}(0) + 4 = 4$, giving point $(0, 4)$
• For x-intercept, let $y = 0$: $0 = -\frac{1}{2}x + 4$, so $\frac{1}{2}x = 4$, therefore $x = 8$, giving point $(8, 0)$

Step 3: State the key characteristics

• Gradient: $-\frac{1}{2}$
• Intercepts: $(0, 4)$ and $(8, 0)$
• Domain: $\{x : x ∈ ℝ\}$
• Range: $\{y : y ∈ ℝ\}$
• Decreasing function: as x increases, y decreases

Worked Example: Quadratic Function

Question: Write the quadratic function $2y - x^2 + 4 = 0$ in standard form. Draw a graph of the function and state the significant characteristics.

Solution:

Step 1: Write the equation in standard form $y = ax^2 + bx + c$

Starting with: $2y - x^2 + 4 = 0$

Rearrange: $2y = x^2 - 4$

Divide by 2: $y = \frac{1}{2}x^2 - 2$

Therefore: $a = \frac{1}{2}$, $b = 0$, $c = -2$

Step 2: Draw a graph of the parabola

Find key points:

• For y-intercept, let $x = 0$: $y = \frac{1}{2}(0)^2 - 2 = -2$, giving point $(0, -2)$
• For x-intercepts, let $y = 0$: $0 = \frac{1}{2}x^2 - 2$, so $\frac{1}{2}x^2 = 2$, therefore $x^2 = 4$, giving $x = ±2$
• This gives us points $(-2, 0)$ and $(2, 0)$

Step 3: State the significant characteristics

• Shape: $a > 0$, therefore the graph is a "smile" (opens upward)
• Intercepts: $(-2, 0)$, $(2, 0)$ and $(0, -2)$
• Turning point: $(0, -2)$
• Axis of symmetry: $x = -\frac{b}{2a} = -\frac{0}{2(\frac{1}{2})} = 0$
• Domain: $\{x : x ∈ ℝ\}$
• Range: $\{y : y ≥ -2, y ∈ ℝ\}$
• The function is decreasing for $x < 0$ and increasing for $x > 0$

Worked Example: Exponential Functions

Question: Draw the graphs of $f(x) = 2^x$ and $g(x) = (\frac{1}{2})^x$ on the same set of axes and compare the two functions.

Solution:

Step 1: Examine the functions and work out the information needed to draw the graphs

For $f(x) = 2^x$:

• If $y = 0$: $2^x = 0$, but $2^x ≠ 0$, so no solution (no x-intercept)
• If $x = 0$: $2^0 = 1$, giving point $(0, 1)$
• Asymptotes: $f(x) = 2^x$ has a horizontal asymptote at $y = 0$ (the x-axis)

For $g(x) = (\frac{1}{2})^x$:

• If $y = 0$: $(\frac{1}{2})^x = 0$, but $(\frac{1}{2})^x ≠ 0$, so no solution (no x-intercept)
• If $x = 0$: $(\frac{1}{2})^0 = 1$, giving point $(0, 1)$
• Asymptotes: $g(x) = (\frac{1}{2})^x$ also has a horizontal asymptote at $y = 0$

Step 2: Draw the graphs of the exponential functions

Step 3: State the significant characteristics

• Symmetry: $f$ and $g$ are symmetrical about the y-axis
• Domain of $f$ and $g$: $\{x : x ∈ ℝ\}$
• Range of $f$ and $g$: $\{y : y > 0, y ∈ ℝ\}$
• Function $g$ decreases as $x$ increases and function $f$ increases as $x$ increases
• The two graphs intersect at the point $(0, 1)$