Sketched below are the graphs of functions $f$ and $g$ defined: $$f(x) = ax^2 + bx + c ext{ and } g(x) = -x - 2$$ - A is the x-intercept of both $f$ and $g$ - NSC Technical Mathematics - Question 4 - 2023 - Paper 1

We know from the given that point R( -3) lies on the straight line $g$:

Substituting $x = -3$ into $g(x)$ gives:

$g(-3) = -(-3) - 2 = 3 - 2 = 1$

Thus, $k = 1$.

The y-intercept of function $f$, which is B, occurs when $x = 0$. Substituting $x = 0$ into $f(x)$:

$f(0) = a(0)^2 + b(0) + c = c$

With values identified earlier, $c = -2$. Therefore, the x-coordinate of B is $0$.

From the information provided, substitute $k = 1$ into $f(x)$:

Using the standard form of quadratic function:

$f(x) = a(x + 2)(x - 4)$

Expanding gives:

$f(x) = ax^2 - 2ax - 8a$

Given that $a = -1$, we substitute for $f(x)$:

$f(x) = -1(x^2 - 2x - 8) = -x^2 + 2x + 8$

To find the range of the function $f(x) = -x^2 + 2x + 8$, we first identify the vertex (maximum point) since it is a downward-facing parabola. The x-value of the vertex is given by

$x = -\frac{b}{2a} = -\frac{2}{2(-1)} = 1$

Substituting $x = 1$ into $f(x)$ gives:

$f(1) = -(1^2) + 2(1) + 8 = 9$

Thus, the range of $f$ is $(-\infty, 9]$.

To find where $f(x) \geq g(x)$, set:

$-x^2 + 2x + 8 \geq -x - 2$

Rearranging gives:

$-x^2 + 3x + 10 \geq 0$

Factoring yields the critical points. Solve to find intervals where this inequality holds.