Question

1. sketch a graph of each polynomial function.

a) $y = (x + 1)(x - 3)(x + 2)$
b) $y = -x(x + 1)(x + 2)^2$
c) $y = (x - 4)^2(x + 3)^3$

Answer:

Part (a): \( y = (x + 1)(x - 3)(x + 2) \)

Step 1: Find the x-intercepts

To find the x-intercepts, set \( y = 0 \):

$$ (x + 1)(x - 3)(x + 2) = 0 $$

This gives the roots \( x = -2 \), \( x = -1 \), and \( x = 3 \). So the x-intercepts are \( (-2, 0) \), \( (-1, 0) \), and \( (3, 0) \).

Step 2: Find the y-intercept

To find the y-intercept, set \( x = 0 \):

$$ y = (0 + 1)(0 - 3)(0 + 2) = (1)(-3)(2) = -6 $$

So the y-intercept is \( (0, -6) \).

Step 3: Determine the end behavior

The leading term of the polynomial (when expanded) is \( x^3 \) (since the product of the leading terms of each factor \( x \cdot x \cdot x = x^3 \)). For a cubic function with a positive leading coefficient (\( a = 1 > 0 \)):

• As \( x \to \infty \), \( y \to \infty \)
• As \( x \to -\infty \), \( y \to -\infty \)

Step 4: Analyze the behavior at each root

• At \( x = -2 \): The multiplicity of the root \( x = -2 \) is 1 (odd), so the graph crosses the x-axis at \( (-2, 0) \).
• At \( x = -1 \): The multiplicity of the root \( x = -1 \) is 1 (odd), so the graph crosses the x-axis at \( (-1, 0) \).
• At \( x = 3 \): The multiplicity of the root \( x = 3 \) is 1 (odd), so the graph crosses the x-axis at \( (3, 0) \).

Step 5: Sketch the graph

1. Plot the x-intercepts \( (-2, 0) \), \( (-1, 0) \), \( (3, 0) \) and the y-intercept \( (0, -6) \).
2. Use the end behavior: as \( x \to \infty \), the graph goes up; as \( x \to -\infty \), the graph goes down.
3. Since all multiplicities are odd, the graph crosses the x-axis at each intercept. Connect the points, making sure to pass through each intercept and follow the end behavior.

Part (b): \( y = -x(x + 1)(x + 2)^2 \)

Step 1: Find the x-intercepts

Set \( y = 0 \):

$$ -x(x + 1)(x + 2)^2 = 0 $$

This gives the roots \( x = 0 \), \( x = -1 \), and \( x = -2 \) (with multiplicity 2). So the x-intercepts are \( (0, 0) \), \( (-1, 0) \), and \( (-2, 0) \) (with multiplicity 2 at \( x = -2 \)).

Step 2: Find the y-intercept

Set \( x = 0 \):

$$ y = -0(0 + 1)(0 + 2)^2 = 0 $$

So the y-intercept is \( (0, 0) \) (which is also an x-intercept).

Step 3: Determine the end behavior

The leading term: expand the polynomial. The leading terms are \( -x \cdot x \cdot x^2 = -x^4 \) (since \( (x + 2)^2 \) has leading term \( x^2 \), \( (x + 1) \) has leading term \( x \), and \( -x \) has leading term \( -x \)). For a quartic function with a negative leading coefficient (\( a = -1 < 0 \)):

• As \( x \to \infty \), \( y \to -\infty \)
• As \( x \to -\infty \), \( y \to -\infty \) (since \( (-x)^4 = x^4 \), and with the negative sign, both ends go down)

Step 4: Analyze the behavior at each root

• At \( x = 0 \): Multiplicity 1 (odd), so the graph crosses the x-axis at \( (0, 0) \).
• At \( x = -1 \): Multiplicity 1 (odd), so the graph crosses the x-axis at \( (-1, 0) \).
• At \( x = -2 \): Multiplicity 2 (even), so the graph touches the x-axis at \( (-2, 0) \) and turns around.

Step 5: Sketch the graph

1. Plot the x-intercepts \( (0, 0) \), \( (-1, 0) \), and \( (-2, 0) \) (with a touch-and-turn at \( (-2, 0) \)).
2. End behavior: both ends go down (since it's a quartic with negative leading coefficient).
3. At \( x = -2 \), the graph touches the axis and turns around (due to even multiplicity). At \( x = 0 \) and \( x = -1 \), the graph crosses the axis (due to odd multiplicity).
4. The leading coefficient is negative, so the graph opens downward (both ends down).

Part (c): \( y = (x - 4)^2(x + 3)^3 \)

Step 1: Find the x-intercepts

Set \( y = 0 \):

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

This gives the roots \( x = 4 \) (multiplicity 2) and \( x = -3 \) (multiplicity 3). So the x-intercepts are \( (4, 0) \) (multiplicity 2) and \( (-3, 0) \) (multiplicity 3).

Step 2: Find the y-intercept

Set \( x = 0 \):

$$ y = (0 - 4)^2(0 + 3)^3 = (16)(27) = 432 $$

So the y-intercept is \( (0, 432) \).

Step 3: Determine the end behavior

The leading term: expand the polynomial. The leading terms are \( x^2 \cdot x^3 = x^5 \) (since \( (x - 4)^2 \) has leading term \( x^2 \) and \( (x + 3)^3 \) has leading term \( x^3 \)). For a fifth-degree polynomial with a positive leading coefficient (\( a = 1 > 0 \)):

• As \( x \to \infty \), \( y \to \infty \)
• As \( x \to -\infty \), \( y \to -\infty \) (since odd degree, positive leading coefficient: right end up, left end down)

Step 4: Analyze the behavior at each root

• At \( x = 4 \): Multiplicity 2 (even), so the graph touches the x-axis at \( (4, 0) \) and turns around.
• At \( x = -3 \): Multiplicity 3 (odd), so the graph crosses the x-axis at \( (-3, 0) \) (with a "flatter" crossing due to higher multiplicity).

Step 5: Sketch the graph

1. Plot the x-intercepts \( (4, 0) \) (touch and turn) and \( (-3, 0) \) (cross, flatter), and the y-intercept \( (0, 432) \).
2. End behavior: as \( x \to \infty \), the graph goes up; as \( x \to -\infty \), the graph goes down.
3. At \( x = 4 \), the graph touches the axis and turns around (even multiplicity). At \( x = -3 \), the graph crosses the axis (odd multiplicity, flatter due to multiplicity 3).
4. Connect the points, following the end behavior and the behavior at each intercept.

Final Answers (Graph Descriptions)

a) The graph of \( y = (x + 1)(x - 3)(x + 2) \) has x-intercepts at \( (-2, 0) \), \( (-1, 0) \), \( (3, 0) \), y-intercept at \( (0, -6) \), crosses the x-axis at each intercept, and has end behavior \( \to \infty \) as \( x \to \infty \) and \( \to -\infty \) as \( x \to -\infty \).

b) The graph of \( y = -x(x + 1)(x + 2)^2 \) has x-intercepts at \( (0, 0) \), \( (-1, 0) \), \( (-2, 0) \) (touches at \( (-2, 0) \)), y-intercept at \( (0, 0) \), crosses the x-axis at \( (0, 0) \) and \( (-1, 0) \), touches at \( (-2, 0) \), and has end behavior \( \to -\infty \) as \( x \to \pm\infty \).

c) The graph of \( y = (x - 4)^2(x + 3)^3 \) has x-intercepts at \( (4, 0) \) (touches) and \( (-3, 0) \) (crosses, flatter), y-intercept at \( (0, 432) \), touches the x-axis at \( (4, 0) \), crosses at \( (-3, 0) \), and has end behavior \( \to \infty \) as \( x \to \infty \) and \( \to -\infty \) as \( x \to -\infty \).