The graphs of the functions $f(x) = -(x + 3)^2 + 4$ and $g(x) = x + 5$ are drawn below - NSC Mathematics - Question 5 - 2023 - Paper 1

The function $f(x)$ represents a downward-opening parabola, which means its range is determined by the turning point and extends downwards to negative infinity. Hence, the range of $f$ is:

$ext{Range of } f: (- ext{∞}, 4].$

To find the x-coordinates where the functions intersect, we set $f(x) = g(x)$:

$-(x + 3)^2 + 4 = x + 5$

Rearranging gives:

$-(x + 3)^2 - x + 4 - 5 = 0$

This simplifies to:

$-(x + 3)^2 - x - 1 = 0$

Multiplying through by -1 leads to:

$(x + 3)^2 + x + 1 = 0$

Expanding gives:

$x^2 + 6x + 9 + x + 1 = 0$ $x^2 + 7x + 10 = 0$

Factoring yields:

$(x + 5)(x + 2) = 0$

Thus, the solutions are $x = -5$ and $x = -2$, confirming the x-coordinates of A and B.

For the equation $-(x + 3)^2 + 4 + c = 0$ to have one negative and one positive root, its discriminant must be greater than or equal to zero. The equation can be rearranged to:

$-(x + 3)^2 + (4 + c) = 0$

This requires setting the vertex (which occurs at $x = -3$) to be at least equal to zero, thus needing:

$(4 + c) > 0 ext{ and } c < 0 ext{ for the roots to have different signs.}$

Therefore, this means:

$c > -4 ext{ and } c < 0.$

Given that $g(x) = x + 5$, we can express $h(x)$ as:

$h(x) = (x + 5) + k.$

Thus, the function $h(x) = x + (5 + k)$. To express it in the requested form, we can define:

$H(x) = h(x) = x + (5 + k).$