Assist me with this graph.

[tex]\begin{aligned}h(x) & = 0\\\implies \dfrac{6}{x-2}+1 & = 0\\\dfrac{6}{x-2} & = -1\\6 & = -1(x-2)\\6 & = -x+2\\4 & = -x\\x & = -4\end{aligned}[/tex]

Therefore, the coordinates of point A are (-4, 0).

The y-intercept of function h(x) is when x = 0:

[tex]\begin{aligned}h(x) & = \dfrac{6}{x-2}+1\\\implies h(0) & = \dfrac{6}{0-2}+1\\ & = -3+1\\& = -2\end{aligned}[/tex]

Therefore, the coordinates of point B are (0, -2).

The x-intercept of function g(x) is when g(x) = 0:

[tex]\begin{aligned}g(x) & = 0\\\implies -2x+8 & = 0\\-2x & = -8\\x & = 4\end{aligned}[/tex]

Therefore, the coordinates of point E are (4, 0).

Question 3.1.2.

As function f(x) passes through point C (-6, 20), substitute the point into the equation of the function to find the value of k:

[tex]\begin{aligned}f(-6) & = 20\\\implies (-6)^2 + k & = 20\\36 + k & = 20\\k & = 20 - 36\\k & = -16\end{aligned}[/tex]

Hence, proving that the value of k is -16.

Question 3.1.3.

The domain is the set of all possible input values (x-values).

The domain of function f(x) is unrestricted, therefore its domain is:

Solution: -∞ < x < ∞
Interval notation: (-∞, ∞)

Question 3.1.4.

The range is the set of all possible output values (y-values).

The range of function f(x) is restricted, since the minimum point of the parabola is at (0, -16). Therefore its range is:

Solution: f(x) ≥ -16
Interval notation: [-16, ∞)

Question 3.1.5.

[tex]\begin{aligned}g(x) - f(x) & \geq 0\\\implies g(x) &\geq 0+f(x)\\ \implies g(x) & \geq f(x)\end{aligned}[/tex]

Therefore, find the interval for which g(x) is greater than or equal to f(x).

From inspection of the given graph this is between points C and E:

Solution: -6 ≤ x ≤ 4
Interval notation: [-6, 4]