The function provided in the question is:

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

Let's go over the questions one by one.

1. At which x-intercept(s) does the graph cross the x-axis?

   • From the image, the correct answer selected is $x = 1, 4, -1$.
   • This means the graph crosses the x-axis at these values of $x$.

2. At which x-intercept(s) does the graph touch the x-axis and turn around?

   • The answer chosen is "There are no x-intercepts at which the graph touches the x-axis and turns around."
   • This is correct because in a cubic function, like the one given, unless there is a double root, the graph usually crosses the x-axis without turning around.

3. Find the y-intercept.

   • The y-intercept occurs when $x = 0$. Plugging $x = 0$ into $f(x)$: $f(0) = 0^3 - 4(0)^2 - 0 + 4 = 4$
   • The y-intercept is $y = 4$, which matches the answer provided in the image.

4. Determine whether the graph has y-axis symmetry, origin symmetry, or neither.

   • The selected answer is currently blank.
   • To check for symmetry:
     • Y-axis symmetry: This happens if $f(-x) = f(x)$, but here $f(-x) \neq f(x)$, so it doesn’t have y-axis symmetry.
     • Origin symmetry: This happens if $f(-x) = -f(x)$. Checking this, we can see it doesn't hold either.
   • Therefore, the correct answer is "neither".

Would you like more details on any specific part?

Here are 5 related questions:

1. How do you find the x-intercepts of a polynomial function?
2. What does it mean if a graph "turns around" at an intercept?
3. How can you algebraically check if a graph has symmetry?
4. How does the degree of a polynomial affect the shape of its graph?
5. Why is the y-intercept always found when $x = 0$?

Tip:

For polynomials, the sign of the leading coefficient and the degree can give you an idea of the end behavior of the graph.