Question

13

(b) (i) on the diagram, sketch the graph of (y = (x + 1)(x - 3)^2).
label the values where the graph meets the (x)-axis and the (y)-axis.

(ii) write ((x + 1)(x - 3)^2) in the form (ax^3 + bx^2 + cx + d).

Explanation:

🆕 New Concept Discovered: Polynomial Curve Sketching
Analyzing roots, multiplicity, and intercepts to draw curves

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Step 1: Find the intercepts

To sketch the graph of \( y = (x+1)(x-3)^2 \), we first find where it crosses or touches the axes.

• \( y \)-intercept: Set \( x = 0 \).

\[ y = (0+1)(0-3)^2 = 1 \cdot 9 = 9 \]
So, the graph crosses the \( y \)-axis at \( (0, 9) \).

• \( x \)-intercepts: Set \( y = 0 \).

\[ (x+1)(x-3)^2 = 0 \]
This gives two roots:

• \( x = -1 \) (with multiplicity 1)
• \( x = 3 \) (with multiplicity 2)

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Step 2: Determine the behavior at the roots and end behavior

• Behavior at \( x = -1 \): Since the factor \( (x+1) \) has an odd power (power of 1), the graph crosses straight through the \( x \)-axis at \( x = -1 \).
• Behavior at \( x = 3 \): Since the factor \( (x-3)^2 \) has an even power (power of 2), the graph is tangent to the \( x \)-axis at \( x = 3 \). This means it touches the axis and turns back around (forming a local minimum).
• End behavior: Expanding the expression gives a leading term of \( x^3 \). Since the degree is odd and the leading coefficient is positive, the graph goes down to negative infinity on the far left (\( x \to -\infty, y \to -\infty \)) and up to positive infinity on the far right (\( x \to \infty, y \to \infty \)).

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Step 3: Sketch the curve

Using the information gathered:

1. Start in the bottom-left quadrant.
2. Go up and cross the \( x \)-axis at \( -1 \).
3. Continue upwards, passing through the \( y \)-axis at \( 9 \), reaching a local maximum peak to the right of the \( y \)-axis.
4. Turn downwards to touch the \( x \)-axis at \( 3 \) (tangent point).
5. Turn back upwards and head towards positive infinity on the right.

(See the final answer section for the visual representation of this sketch.)

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Step 4: Expand the expression for part (ii)

We need to write \( (x+1)(x-3)^2 \) in the expanded form \( ax^3 + bx^2 + cx + d \).

First, expand the squared term:
\[ (x-3)^2 = x^2 - 6x + 9 \]

Next, multiply this result by \( (x+1) \):
\[ (x+1)(x^2 - 6x + 9) = x(x^2 - 6x + 9) + 1(x^2 - 6x + 9) \]
\[ = x^3 - 6x^2 + 9x + x^2 - 6x + 9 \]

Now, group and combine the like terms:
\[ = x^3 + (-6x^2 + x^2) + (9x - 6x) + 9 \]
\[ = x^3 - 5x^2 + 3x + 9 \]

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Answer:

(b) (i) Sketch of the graph:

Below is the sketch of the curve \( y = (x+1)(x-3)^2 \):

          y
          ^
          |          * (local max)
          |        *   *
        9 +      *       *
          |    *           *
          |  *               *
   -------*-------------------*-------> x
        -1|                  3
          |

• Key features to label on your diagram:
• The curve crosses the \( x \)-axis at \( -1 \).
• The curve crosses the \( y \)-axis at \( 9 \).
• The curve touches and turns around (is tangent to) the \( x \)-axis at \( 3 \).

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(b) (ii) Expanded form:

\[ x^3 - 5x^2 + 3x + 9 \]