For the function $y=x^{4}-7 x^{3}-8 x^{2}$, its $y$-intercept is $y=$ its $x$-intercepts are $x=$
Note: If there is more than one answer enter them separated by commas. If there are none, enter none .
When $x \rightarrow \infty, y \rightarrow$
$\infty$ (Input + or - for the answer)
When $x \rightarrow-\infty, y \rightarrow$
$\infty$ (Input + or - for the answer)

Step 1 ：The y-intercept of a function is the point where the function crosses the y-axis. This happens when x = 0. So, to find the y-intercept, we substitute x = 0 into the function, which gives us $y=0$.

Step 2 ：The x-intercepts of a function are the points where the function crosses the x-axis. This happens when y = 0. So, to find the x-intercepts, we set the function equal to zero and solve for x, which gives us $x=-1, 0, 8$.

Step 3 ：As x approaches infinity, we look at the highest power of x in the function. Since the coefficient of x^4 is positive, as x approaches infinity, y will also approach infinity.

Step 4 ：Similarly, as x approaches negative infinity, we look at the highest power of x in the function. Since the highest power of x is even, y will also approach infinity, regardless of whether x is positive or negative.

Step 5 ：Final Answer: The y-intercept is \(\boxed{y=0}\). The x-intercepts are \(\boxed{x=-1, 0, 8}\). As \(x \rightarrow \infty, y \rightarrow +\infty\). As \(x \rightarrow-\infty, y \rightarrow +\infty\).