Sketched below are the graphs of $f(x)=-2x^2 + 4x + 16$ and $g(x)=2x + 4$ - NSC Mathematics - Question 5 - 2021 - Paper 1

The turning point of a quadratic function can be found using the formula for the x-coordinate:

$x = -\frac{b}{2a}$,

where $a = -2$ and $b = 4$.

Substituting these values:

$x = -\frac{4}{2 \times -2} = 1$

Now, substitute $x = 1$ back into the function to find the y-coordinate:

$f(1) = -2(1)^2 + 4(1) + 16 = -2 + 4 + 16 = 18$

Thus, the coordinates of C are (1, 18).

Since the leading coefficient of $f(x)$ is negative, the parabola opens downwards. The maximum value occurs at the turning point. Therefore, the range of $f$ is:

$y \leq 18$

Thus, the range is: $(-\infty, 18]$.

The maximum value of $h(x)$ at $x = 2$ means:

$h(2) = f(2+p) + q = 15$

First, we find $f(2) = -2(2)^2 + 4(2) + 16 = -8 + 8 + 16 = 16$.

Thus, we can write:

$f(2+p) + q = 15$

Since we know $f(2) = 16$, we have:

$16 + q = 15 \Rightarrow q = -1$

Now, for the turning point to occur at $x = 2$, we should have:

$2 + p = 1 \Rightarrow p = -1$.

Therefore, $p = -1$ and $q = -1$.

To find the inverse, we start with:

$y = g(x) = 2x + 4$

Now, we will swap x and y:

$x = 2y + 4$

Next, solve for y:

$x - 4 = 2y \Rightarrow y = \frac{x - 4}{2}$

Thus, the inverse function is:

$g^{-1}(x) = \frac{x - 4}{2}.$

To find where the inverse function equals zero:

$g^{-1}(x) = 0 \Rightarrow \frac{x - 4}{2} = 0$

Multiplying both sides by 2 gives:

$x - 4 = 0 \Rightarrow x = 4.$

For the functions to NOT intersect, the equation:

$f(x) + k = g(x)$

needs to have no real solutions. Setting both functions to equal gives us:

$-2x^2 + 4x + 16 + k = 2x + 4$

Rearranging:

$-2x^2 + 2x + (12 + k) = 0$

This is a quadratic equation, and for it to have no solutions, the discriminant must be less than zero:

$b^2 - 4ac < 0 \Rightarrow 2^2 - 4(-2)(12+k) < 0$

Solving gives:

$4 + 8(12 + k) < 0 \Rightarrow 100 + 8k < 0 \Rightarrow k < -12.5$

Therefore, the values of k for which p and g will NOT intersect are:

$k < -12.5$.