4.1 Write down: (a) The range of $f$ - NSC Technical Mathematics - Question 4 - 2022 - Paper 1

Point $Q$ is the reflection of point $P(0, -5)$ about the line $x = 2$. The x-coordinate of $Q$ will be:

$Q_x = 2 + (2 - 0) = 4.$

The y-coordinate remains the same as point $P$, so:

$Q = (4, -5).$

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

$-x^2 + 4x - 5 = 0.$
Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{4^2 - 4(-1)(-5)}}{2(-1)} = \frac{-4 \pm \sqrt{16 - 20}}{-2}.$
Thus:

$x = 4.$
The only x-intercept is at $x = 4$.

The length of segment $AB$ can be calculated using the coordinates of points $A$ and $B$. With $A(4, 0)$ and $B(0, -5)$, the distance formula gives:

$AB = \sqrt{(4 - 0)^2 + (0 - (-5))^2} = \sqrt{16 + 25} = \sqrt{41}.$

The slope-intercept form of $g(x)$ is given as $g(x) = mx + c$. Since point $A(-1, 0)$ lies on $g$, we can find $c$ using $g(-1) = 0$:
Setting $0 = -m + c$... We also need point $P(0, -5)$:
For this point, $-5 = c$.
From $g(-1) = -5$, we have:

$c = -5, m = -1.$

Analyzing the sign of the product $f(x) \times g(x)$, we need to find intervals where both functions are positive or both negative. We found:

• For $f(x)$: Positive between $(-\infty, 0)$ and $(0, 2)$, and negative afterward.
• For $g(x)$: Positive when $x > -5$, negative otherwise.

Thus, the values of $x$ are within the intervals $(-\infty, -5)$ and approximately $(0, 2)$ where the product is positive.