Part 1: End Behavior of the Graph

To analyze the end behavior of the function $f(x) = x^3 - 4x^2 - x + 4$, we examine the degree and leading coefficient of the polynomial.

• The degree of the polynomial is 3 (since the highest exponent of $x$ is 3).
• The leading coefficient is 1 (the coefficient of $x^3$).

For polynomials with odd degrees and positive leading coefficients, the graph will:

• Rise to the right (positive direction as $x \to +\infty$).
• Fall to the left (negative direction as $x \to -\infty$).

Thus, the correct answer is: C. The graph falls to the left and rises to the right.

Part 2: Finding the x-intercepts

To find the x-intercepts of the function, we set $f(x) = 0$ and solve for $x$:

$x^3 - 4x^2 - x + 4 = 0$

Using factoring or other methods, we find the roots (x-intercepts): $x = 1, \, x = 4, \, x = -1$

Thus, the x-intercepts are: $\boxed{1, 4, -1}$

Part 3: Crossing the x-axis at the x-intercepts

Now, we determine whether the graph crosses the x-axis or touches and turns around at each intercept.

• To do this, we analyze the multiplicity of the roots.
• Since the polynomial is of degree 3, and all roots have multiplicity 1, the graph crosses the x-axis at each intercept.

Therefore, the graph crosses the x-axis at the following points: $\boxed{1, 4, -1}$

Answer: A. $x = 1, 4, -1$

Part 4: Touching the x-axis and turning around

For the graph to touch the x-axis and turn around at an intercept, the multiplicity of the root must be even. Since all of the roots of this polynomial have multiplicity 1 (odd), the graph does not touch and turn around at any intercept.

Thus, the answer is: B. There are no x-intercepts at which the graph touches the x-axis and turns around.

Part 5: Finding the y-intercept

To find the y-intercept, we substitute $x = 0$ into the function:

$f(0) = 0^3 - 4(0)^2 - (0) + 4 = 4$

Thus, the y-intercept is $y = 4$.

Answer: $y = 4$

Part 6: Symmetry of the Graph

To determine whether the graph has symmetry, we check:

1. y-axis symmetry: A graph has y-axis symmetry if $f(-x) = f(x)$.
2. origin symmetry: A graph has origin symmetry if $f(-x) = -f(x)$.

For this function, we compute $f(-x)$:

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

Since $f(-x) \neq f(x)$ and $f(-x) \neq -f(x)$, the graph has neither y-axis symmetry nor origin symmetry.

Thus, the answer is: C. Neither

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Let me know if you'd like more details or if you have further questions on this! Here's a tip to help with polynomials:

Tip:

When determining the end behavior of a polynomial, focus on the degree and leading coefficient. If the degree is odd and the leading coefficient is positive, the graph will fall to the left and rise to the right. If the leading coefficient is negative, the graph will fall to the right and rise to the left.