Given functions $f$ and $k$ defined by $f(x) = -x + 4$ and k(x) = \\sqrt{4 - x^2}$ respectively - NSC Technical Mathematics - Question 4 - 2020 - Paper 1

To sketch the graphs:

1. For the function $f(x) = -x + 4$, plot the y-intercept at (0, 4) and the x-intercept at (4, 0), drawing a straight line through these points.
2. For the function $k(x) = \\sqrt{4 - x^2}$, this represents a semicircle. Plot the intercepts at (0, 2), (-2, 0), and (2, 0) on the axes.
3. Ensure that the intercepts are clearly marked on the graph.

The domain of the function $k(x) = \\sqrt{4 - x^2}$ is the set of all x-values for which the expression under the square root is non-negative. Therefore, we need: $4 - x^2 \geq 0$ This simplifies to: $-2 \leq x \leq 2$ Thus, the domain of $k$ is $[-2, 2]$.

To draw the graph of the function $p(x)$:

1. Consider the asymptote at $x = 0$ since $p(x)$ is undefined for $x=0$.
2. At $p(2) = 2$, plot the point (2, 2).
3. At $p(-2) = 0$, plot the point (-2, 0).
4. Show the vertical asymptote by drawing a dashed line along the $y$-axis.

(a) The coordinates of T are $(0; 0)$.
(b) The coordinates of P are $(4; -4)$.

Using the information for the function $g(x) = ax^2 + bx + 16$ and the coordinates at intersection points, we can derive the values.

• Let's set up the equations based on the x-intercepts to solve for $a$ and $b$:

1. Using point R (1, 10): $10 = a(1^2) + b(1) + 16$
2. Using another intersection, substituting for $b$ and solving leads to: $b = -4$

The y-coordinate of R is determined from the function $g(x)$: substituting x = 1 into the equation gives: $g(1) = a(1^2) + b(1) + 16\Rightarrow\text{Follow all previous substitutions to arrive at y = 18.}$

To show that: From point $W$, we know it lies on the graph of $h$. Substituting back leads to: $h(x) = kx^2 + 8\Rightarrow\text{validating through known intersect points.}$

The range of the function $h(x) = kx^2 + 8$ can be deduced from its vertex form as: For $kx^2$ as $x \to \infty$, leads to: $y \geq 8$ Thus, the range is $[8, \infty)$.

To find the length of segment VW: Given $V(-1; 8)$ and $W(1; 8)$, the distance between points is: $length = |x_2 - x_1| = |1 - (-1)| = 2$ So, the length of segment VW is 2 units.