Given: Functions $f$ and $h$ defined by $f(x) = -2(x-3)^2 + 18$ and h(x) = 2x + c 4.1.1 Write down the coordinates of the turning point of $f$ - NSC Technical Mathematics - Question 4 - 2024 - Paper 1

To find the x-intercepts, set $f(x) = 0$:

In the sketch, plot the turning point at $(3, 18)$ and the x-intercepts at $(0, 0)$ and $(6, 0)$. The graph is a downward opening parabola.

To find $t$, substitute $x = 5$ into the function $f$:

$f(5) = -2(5-3)^2 + 18 = -2(2^2) + 18 = -8 + 18 = 10$

Thus, $t = 10$.

Use the point of intersection $(5, 10)$ with $h(x) = 2x + c$:

The line defined by $h(x) = 2x$ intersects the y-axis at $(0, 0)$ and the point $(5, 10)$, which is already plotted. The line is linear and will rise steeply.

The domain of p(x) = -rac{8}{x} is all real numbers except for $x = 0$, thus:

The function $g(y) = a^2 + q$ suggests that the range is dependent on the value of $a^2$ and $q$. Assuming $a^2 ightarrow ext{all non-negative values}$, the range for $g$ is:

$R : [q, ext{∞})$

Given that the horizontal asymptote is $y = 0$, we can set $q = -4$ so that the function $g$ approaches this asymptote.

To find the coordinates of $D$, evaluate the function $p$ at $x=1$:

p(1) = -rac{8}{1} = -8

Therefore, the coordinates of $D$ are $(0, -8)$.

To find $C$, we use the function graph. The coordinates will depend on the intersection points and will be $(2, 0)$ due to the characteristics of the graph.

Using given function characteristics, solve for $a$ such that $g(-2) = 4$ yields a = rac{1}{2}.