Question One Carefully sketch the graph of the cubic function $$ f(x)=x^{3}+3 x^{2}-x-3 $$ clearl showing the intercepts and vertices $$ [6 \mathrm{mks}] $$

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**Step 1. Find the intercepts**  
- **$$ y $$-intercept:**  
  Set $$ x=0 $$ in  
  $$
  f(x)=x^3+3x^2-x-3.
  $$  
  Then,  
  $$
  f(0)=0+0-0-3=-3.
  $$  
  So the $$ y $$-intercept is at $$(0,-3)$$.

- **$$ x $$-intercepts:**  
  Solve  
  $$
  x^3+3x^2-x-3=0.
  $$  
  Notice that $$ f(1)=1^3+3(1)^2-1-3=1+3-1-3=0 $$; hence, $$ x=1 $$ is a root.  
  Factorizing by dividing by $$ (x-1) $$, we get
  $$
  x^3+3x^2-x-3 = (x-1)(x^2+4x+3).
  $$  
  Next, factor the quadratic  
  $$
  x^2+4x+3=(x+1)(x+3).
  $$  
  Therefore, the factorization is
  $$
  f(x)=(x-1)(x+1)(x+3)=0.
  $$
  The zeros are
  $$
  x=1,\quad x=-1,\quad x=-3.
  $$

**Step 2. Calculate the critical points (vertices)**  
To find the local maximum and minimum, compute the derivative:
$$
f'(x)=\frac{d}{dx}\left(x^3+3x^2-x-3\right)=3x^2+6x-1.
$$  
Set $$ f'(x)=0 $$:
$$
3x^2+6x-1=0.
$$
Apply the quadratic formula:
$$
x=\frac{-6\pm\sqrt{6^2-4\cdot3\cdot(-1)}}{2\cdot3}=\frac{-6\pm\sqrt{36+12}}{6}=\frac{-6\pm\sqrt{48}}{6}.
$$
Since $$\sqrt{48}=4\sqrt{3}$$, we have
$$
x=\frac{-6\pm4\sqrt{3}}{6}=-1\pm\frac{2\sqrt{3}}{3}.
$$
Thus the critical points are  
$$
x_1=-1+\frac{2\sqrt{3}}{3}\quad \text{and}\quad x_2=-1-\frac{2\sqrt{3}}{3}.
$$

Let’s approximate these values:
- $$ x_1\approx -1+1.155=0.155 $$  
- $$ x_2\approx -1-1.155=-2.155 $$

Now compute the function values:

- For $$ x_1\approx0.155 $$:
  $$
  f(0.155)\approx (0.155)^3+3(0.155)^2-0.155-3.
  $$
  Approximations:
  $$
  (0.155)^3\approx0.0037,\quad 3(0.155)^2\approx3(0.0240)=0.072,\quad -0.155,\quad -3.
  $$
  Therefore,
  $$
  f(0.155)\approx 0.0037+0.072-0.155-3\approx -3.079.
  $$

- For $$ x_2\approx -2.155 $$:
  $$
  f(-2.155)\approx (-2.155)^3+3(-2.155)^2-(-2.155)-3.
  $$
  Approximations:
  $$
  (-2.155)^3\approx -10.06,\quad (-2.155)^2\approx4.647,\quad 3(4.647)\approx13.941,\quad +2.155,\quad -3.
  $$
  Thus,
  $$
  f(-2.155)\approx -10.06+13.941+2.155-3\approx 2.036.
  $$

Determine the nature (using the second derivative). Compute
$$
f''(x)=\frac{d}{dx}(3x^2+6x-1)=6x+6.
$$
- At $$ x_1\approx0.155 $$:
  $$
  f''(0.155)=6(0.155)+6\approx0.93+6=6.93>0,
  $$
  hence $$ (0.155, -3.079) $$ is a local minimum.

- At $$ x_2\approx -2.155 $$:
  $$
  f''(-2.155)=6(-2.155)+6\approx -12.93+6=-6.93<0,
  $$
  so $$ (-2.155, 2.036) $$ is a local maximum.

**Step 3. Determine the point of inflection**  
Find the inflection point by setting the second derivative equal to 0:
$$
6x+6=0\quad\Longrightarrow\quad x=-1.
$$
Compute $$ f(-1) $$:
$$
f(-1)=(-1)^3+3(-1)^2-(-1)-3=-1+3+1-3=0.
$$
Thus, the inflection point is at $$(-1,0)$$.

**Step 4. Summarize key points for the sketch**

- **$$ x $$-intercepts:**  
  $$(1,0)$$, $$(-1,0)$$, and $$(-3,0)$$.

- **$$ y $$-intercept:**  
  $$(0,-3)$$.

- **Local extrema:**  
  - Local maximum at approximately $$(-2.155,\,2.036)$$.  
  - Local minimum at approximately $$(0.155,\,-3.079)$$.

- **Inflection point:**  
  $$(-1,0)$$.

- **End behavior:**  
  As $$ x\to\infty $$, $$ f(x)\to\infty $$; as $$ x\to-\infty $$, $$ f(x)\to-\infty $$.

**Step 5. Sketching guidelines**  
- Plot the intercepts:  
  - $$ x $$-intercepts at $$(-3,0)$$, $$(-1,0)$$, and $$(1,0)$$.  
  - $$ y $$-intercept at $$(0,-3)$$.
- Mark the local maximum at approximately $$(-2.155,2.036)$$ and the local minimum at approximately $$(0.155,-3.079)$$.
- Identify the inflection point at $$(-1,0)$$, where the curve changes concavity.
- Draw the curve starting from the lower left approaching $$-\infty$$, rising towards the local maximum, passing through the intercepts and the inflection point, dropping to the local minimum, and finally increasing towards $$\infty$$ as $$ x\to\infty $$.

Following these steps will produce a clear sketch of the cubic function $$ f(x)=x^3+3x^2-x-3 $$ with all required intercepts and significant points highlighted.
To sketch the graph of $$ f(x) = x^3 + 3x^2 - x - 3 $$: 1. **Intercepts:** - **$$ y $$-intercept:** $$(0, -3)$$ - **$$ x $$-intercepts:** $$(-3, 0)$$, $$(-1, 0)$$, and $$(1, 0)$$ 2. **Local Extrema:** - **Local Maximum:** Approximately at $$(-2.155, 2.036)$$ - **Local Minimum:** Approximately at $$(0.155, -3.079)$$ 3. **Inflection Point:** $$(-1, 0)$$ 4. **End Behavior:** - As $$ x \to \infty $$, $$ f(x) \to \infty $$ - As $$ x \to -\infty $$, $$ f(x) \to -\infty $$ **Sketching Guidelines:** - Plot the intercepts and extrema. - Draw the curve starting from the lower left, passing through the intercepts and inflection point, reaching the local maximum, then the local minimum, and finally increasing towards the upper right. This will create a clear and accurate sketch of the cubic function with all necessary features.

Question One Carefully sketch the graph of the cubic function \( f(x)=x^{3}+3 x^{2}-x-3 \) clearl showing the intercepts and vertices \( [6 \mathrm{mks}] \)

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