Given: $$g(x) = 2^{x - 1}$$ and $$h(x) = -\frac{6}{x} - 1$$ 4.1.1 Write down the equations of the asymptotes of $$h$$ - NSC Technical Mathematics - Question 4 - 2018 - Paper 1

To find the x-intercept, set $h(x) = 0$:

$0 = -\frac{6}{x} - 1$

Solving for $x$:

$\frac{6}{x} = -1 \\ 6 = -x \\ x = -6$

Thus, the x-intercept is at the point $( -6; 0 )$.

Sketch both functions on the axes considering the asymptotes previously established. Ensure to label the asymptotes and intercepts:

• The graph of $g(x) = 2^{x - 1}$ will show exponential growth.
• The graph of $h(x) = -\frac{6}{x} - 1$ will approach its asymptotes accordingly.

To verify that the point $( -2 ; 3 )$ is on the graph of $g$, substitute $x = -2$ into the function:

$g(-2) = 2^{-2 - 1} = 2^{-3} = \frac{1}{8}$.

The provided point does not satisfy the equation, therefore $( -2; 3 )$ is not a point on the graph.

The range of the exponential function $g(x) = 2^{x-1}$ is:

$y > 0$, since the function increases indefinitely but never reaches zero.

The domain of the function $h(x) = -\frac{6}{x} - 1$ is all real numbers except for the value that makes the denominator zero, which is:

$x \neq 0$.

The coordinates of point M, which is the midpoint of TM, can be expressed as:

$M( 1; 0 )$.

To find the length of segment TR, use the difference between y-coordinates:

Let MT = 8 and MR = $g(1) = g(0)$. Calculate:

$TR = MT - MR = 8 - \sqrt{35}$.

To find the intercepts of $f(x) = a(x + p)^2 + q$, substitute $x = 0$: $f(0) = a(0 + p)^2 + q = 6.$ Therefore, we conclude:

• The intercepts are at $(0 ; 6)$.

Expanding the equation:

Starting with $(x - 1)(x - 3) = x^2 - 4x + 3$, multiply by $-2$:

$f(x) = -2(x^2 - 4x + 3) = -2x^2 + 8x - 6.$

Thus, the given form is confirmed.