4.1 Given: $h(x) = \left(\frac{1}{4}\right)(x - 5)$ and $k(x) = -2x^2 + 4x - 6$ 4.1.1 Determine: (a) The y-intercept of $h$ (b) The equation of the asymptote of $h$ (c) The x- and y-intercept(s) of $k$ (d) The turning point of $k$ - NSC Technical Mathematics - Question 4 - 2021 - Paper 1

The function $h(x) = \left(\frac{1}{4}\right)(x - 5)$ is a linear function, and linear functions do not have vertical asymptotes. Since there is no restriction on $x$, the asymptote can be considered as $y = -1.25$ (the y-intercept). Thus, the equation of the asymptote is:

$y = \lim_{x \to \pm \infty} h(x) = -1.25.$

To find the y-intercept of $k$, we set $x = 0$:

$k(0) = -2(0)^2 + 4(0) - 6 = -6.$

So, the y-intercept is $(0, -6)$.

To find the x-intercepts, we solve the equation $k(x) = 0$:

$-2x^2 + 4x - 6 = 0$

This simplifies to:

$x^2 - 2x + 3 = 0.$

Applying the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(3)}}{2(1)}.$

This results in complex roots, so no real x-intercepts exist.

To find the turning point of $k(x) = -2x^2 + 4x - 6$, we can use the vertex formula:

$x = -\frac{b}{2a} = -\frac{4}{2(-2)} = 1.$

Substituting $x = 1$ back into $k(x)$ gives:

$k(1) = -2(1)^2 + 4(1) - 6 = 8.$

Thus, the turning point is at the point $(1, 8)$.

The coordinates of point C, the intersection of $f$ and $g$, is where both equations equal each other. Solving (-\sqrt{-x^2 - x} = 2x - 10), we find:

By substituting $x = 3$, we can compute that $C(3, -4)$ is indeed the intersection point.

To find the length of segment DB, we need to find the coordinates of B and D.
For B (the y-intercept of $f$), when $x = 0$:

$f(0) = -\sqrt{-0^2 - 0} = 0.$
So, B(0, -10).
For D (the y-intercept of $g$), we can see D(0, -10).
Thus, the distance DB is given by:

Distance formula:
$ext{Length} (DB) = \sqrt{(0 - 0)^2 + (-10 + 10)^2} = 5.$

The equation of $f$ is given as:

$f(x) = -\sqrt{-x^2 - x}.$

To find when $g(x) - f(x) > 0$, we set up the inequality:

$2x - 10 - (-\sqrt{-x^2 - x}) > 0$

This simplifies to finding the values of $x$ for which this inequality holds true, requiring further algebraic manipulation to find critical values and intervals.