Given the function of f(x)=6−2x2, Complete the table in ADDENDUM A by determining y-values for the corresponding x-values.

The value of $p$ is $\boxed{\mathbf{25}}$.

1.6.2 Determine the $n^{th}$ term of the number pattern.

The $n^{th}$ term is $\boxed{\mathbf{T_n = 41 - 4n}}$.

1.6.3 Determine the $15^{th}$ term of the sequence.

The $15^{th}$ term is $\boxed{\mathbf{-19}}$.

QUESTION 2

2.1 Given the function of $f(x)=6-2x^2$

2.1.1 Complete the table in ADDENDUM A by determining y-values for the corresponding x-values.
The ADDENDUM A table is not provided in the image. To complete such a table, you would substitute the given $x$-values into the function $f(x)=6-2x^2$ to find the corresponding $y$-values.
For example:

• If $x = -2$, $f(-2) = 6 - 2(-2)^2 = 6 - 2(4) = 6 - 8 = -2$.
• If $x = 0$, $f(0) = 6 - 2(0)^2 = 6 - 0 = 6$.
• If $x = 2$, $f(2) = 6 - 2(2)^2 = 6 - 2(4) = 6 - 8 = -2$.

2.1.2 Sketch the graph of $f(x)=6-2x^2$ in ADDENDUM A showing all intercepts.
The ADDENDUM A graph paper is not provided. Here are the key features for sketching the graph:

• This is a parabola that opens downwards because the coefficient of $x^2$ is negative ($-2$).
• Y-intercept: Set $x=0$.

The y-intercept is $\boxed{\mathbf{(0, 6)}}$.

• X-intercepts: Set $f(x)=0$.

The x-intercepts are $\boxed{\mathbf{(\sqrt{3}, 0) \text{ and } (-\sqrt{3}, 0)}}$.

• Turning Point: For a parabola $f(x)=ax^2+bx+c$, the x-coordinate of the turning point is $x = -\frac{b}{2a}$. Here $a=-2$ and $b=0$.

The y-coordinate is $f(0)=6$. The turning point is $\boxed{\mathbf{(0, 6)}}$.

2.1.3 What is the mathematical name of the graph of $f(x)$?
The graph of $f(x)=6-2x^2$ is a $\boxed{\mathbf{parabola}}$.

2.1.4 Is the graph a function or a non-function? Give a reason for your answer.
The graph is a $\boxed{\mathbf{function}}$.
Reason: For every input value of $x$, there is exactly one output value of $y$. This can be verified by the vertical line test, where any vertical line drawn through the graph will intersect it at most once.

2.1.5 Write down the coordinates of the turning point of $f(x)$.
The coordinates of the turning point are $\boxed{\mathbf{(0, 6)}}$.

2.2 The diagram below represents a graph of $f(x)=ax+q$.

2.2.1 Determine the equation for $f(x)$.

The y-intercept is $(0, 2)$. So, $q=2$.
Another clear point is $(-2, 0)$ (the x-intercept).

The equation for $f(x)$ is $\boxed{\mathbf{f(x) = x+2}}$.

2.2.2 Is the gradient of this graph negative or positive?
The gradient of this graph is $\boxed{\mathbf{positive}}$.
This is because the line slopes upwards from left to right, and the calculated gradient is $a=1$, which is a positive value.