Answer:
[tex]\textsf{The function $y=-\sqrt[3]{x-4}-1$ is always $\boxed{\sf decreasing}$ with the right end}\\\\ \textsf{behaviour of $\boxed{\textsf{as}\;x\rightarrow \infty, y \rightarrow -\infty}$ and an $x$-intercept of $\boxed{(3, 0)}$\:.}[/tex]
Step-by-step explanation:
Given cube root function:
[tex]y=-\sqrt[3]{x-4}-1[/tex]
As x becomes larger and larger, x - 4 increases without bound. The cube root of a number increases as the number increases. Since the cube root part of the function is negative, and the subtraction of 1 further shifts the entire graph downward, the function as a whole is always decreasing as x increases.
The x-intercept is the point at which a graph intersects or crosses the x-axis, and its coordinates have the form (x, 0). Therefore, to find the x-intercept of the given function, set the function equal to zero and solve for x:
[tex]\begin{aligned}-\sqrt[3]{x-4}-1&=0\\\\-\sqrt[3]{x-4}&=1\\\\\sqrt[3]{x-4}&=-1\\\\(\sqrt[3]{x-4})^3&=(-1)^3\\\\x-4&=-1\\\\x&=3\end{aligned}[/tex]
Therefore, the x-intercept is (3, 0).
The right end behavior of a function describes what happens to the function's values as the x-value approaches positive infinity. As we have already established that the function is always decreasing and crosses the x-axis at x = 3, the function will approach negative infinity as x approaches positive infinity:
