5.1 Draw the graph of the following in the ANSWER BOOK, showing all the intercepts on the axes: $$ 9 x^{2}+4 y^{2}=36 $$ 5.2 Determine $$ \frac{d y}{d x} $$ of the function $$ y=\frac{2}{x^{2}}+\sqrt{x}+\frac{1}{x} \pi $$ by making use of the rules of differentiation. The final answer must be with positive exponents and in surd form where applicable. Given: $$ f(x)=-2 x^{3}-x^{2} $$ $$ \begin{array}{l} ext { If }(0 ; 0) ext { is one of the turning points of the graph } f(x) ext {, make use of } \ ext { differentiation to determine the coordinates of the other turning point of the } \ ext { graph of } f(x) ext {. } \ ext { (a;0) and }(0 ; 0) ext { are the points of intersection of } f ext { with the x-axis. } \ ext { D.3.3 Determine the value of } a ext {. } \ ext { Sketch the graph of } f(x) ext {, showing all the points of intercepts on the axes } \ ext { and the turning points. }\end{array} $$

Question

5.1 Draw the graph of the following in the ANSWER BOOK, showing all the intercepts on the axes: $$ 9 x^{2}+4 y^{2}=36 $$ 5.2 Determine $$ \frac{d y}{d x} $$ of the function $$ y=\frac{2}{x^{2}}+\sqrt{x}+\frac{1}{x} \pi $$ by making use of the rules of differentiation. The final answer must be with positive exponents and in surd form where applicable. Given: $$ f(x)=-2 x^{3}-x^{2} $$ $$ \begin{array}{l}	ext { If }(0 ; 0) 	ext { is one of the turning points of the graph } f(x) 	ext {, make use of } \ 	ext { differentiation to determine the coordinates of the other turning point of the } \ 	ext { graph of } f(x) 	ext {. } \ 	ext { (a;0) and }(0 ; 0) 	ext { are the points of intersection of } f 	ext { with the x-axis. } \ 	ext { D.3.3 Determine the value of } a 	ext {. } \ 	ext { Sketch the graph of } f(x) 	ext {, showing all the points of intercepts on the axes } \ 	ext { and the turning points. }\end{array} $$

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**5.1 Graph of the ellipse $$9x^2+4y^2=36$$**

1. Rewrite the equation in standard form by dividing by 36:  
   $$
   \frac{9x^2}{36}+\frac{4y^2}{36}=1 \quad \Longrightarrow \quad \frac{x^2}{4}+\frac{y^2}{9}=1
   $$

2. Identify the intercepts:  
   - **x-intercepts:** Set $$y=0$$ in $$\frac{x^2}{4}+\frac{0^2}{9}=1$$, giving  
     $$
     \frac{x^2}{4}=1 \quad \Longrightarrow \quad x^2=4 \quad \Longrightarrow \quad x=\pm 2.
     $$
   - **y-intercepts:** Set $$x=0$$ in $$\frac{0^2}{4}+\frac{y^2}{9}=1$$, giving  
     $$
     \frac{y^2}{9}=1 \quad \Longrightarrow \quad y^2=9 \quad \Longrightarrow \quad y=\pm 3.
     $$

3. The ellipse has center at $$(0,0)$$ with horizontal semi-axis $$2$$ and vertical semi-axis $$3$$. Plot the points $$(\pm 2,0)$$ and $$(0,\pm 3)$$ and sketch a smooth, elongated circle (ellipse) passing through these intercepts.

---

**5.2 Differentiation of $$y=\frac{2}{x^{2}}+\sqrt{x}+\frac{1}{x}\pi$$**

1. Express each term with exponents:  
   $$
   y=2x^{-2}+x^{\frac{1}{2}}+\pi x^{-1}
   $$

2. Differentiate term by term using the power rule:
   - Differentiate $$2x^{-2}$$:  
     $$
     \frac{d}{dx}\left(2x^{-2}\right)=2\cdot(-2)x^{-3}=-4x^{-3}
     $$
   - Differentiate $$x^{\frac{1}{2}}$$:  
     $$
     \frac{d}{dx}\left(x^{\frac{1}{2}}\right)=\frac{1}{2}x^{-\frac{1}{2}}
     $$
   - Differentiate $$\pi x^{-1}$$:  
     $$
     \frac{d}{dx}\left(\pi x^{-1}\right)=\pi\cdot(-1)x^{-2}=-\pi x^{-2}
     $$

3. Combine the derivatives:  
   $$
   \frac{dy}{dx}=-4x^{-3}+\frac{1}{2}x^{-\frac{1}{2}}-\pi x^{-2}
   $$

4. Express the derivatives with positive exponents:  
   $$
   \frac{dy}{dx}=-\frac{4}{x^3}+\frac{1}{2\sqrt{x}}-\frac{\pi}{x^2}
   $$

---

**D.3.3 Analysis of $$f(x)=-2x^3-x^2$$**

1. **Finding the turning points:**

- Compute the derivative:  
     $$
     f'(x)=\frac{d}{dx}\left(-2x^3-x^2\right)=-6x^2-2x
     $$
     
   - Set $$f'(x)=0$$:  
     $$
     -6x^2-2x=0 \quad \Longrightarrow \quad -2x(3x+1)=0
     $$
     
   - Solve for $$x$$:  
     $$
     -2x=0 \quad \Longrightarrow \quad x=0, \qquad 3x+1=0 \quad \Longrightarrow \quad x=-\frac{1}{3}.
     $$
     
   - Since $$(0,0)$$ is given as one turning point, the other turning point is for $$x=-\frac{1}{3}$$. Find its $$y$$-coordinate:  
     $$
     f\left(-\frac{1}{3}\right)=-2\left(-\frac{1}{3}\right)^3-\left(-\frac{1}{3}\right)^2
     $$
     
     Calculate each term:  
     $$
     \left(-\frac{1}{3}\right)^3=-\frac{1}{27} \quad \Longrightarrow \quad -2\left(-\frac{1}{27}\right)=\frac{2}{27},
     $$
     $$
     \left(-\frac{1}{3}\right)^2=\frac{1}{9} \quad \Longrightarrow \quad -\frac{1}{9}.
     $$
     
     Thus,  
     $$
     f\left(-\frac{1}{3}\right)=\frac{2}{27}-\frac{1}{9}=\frac{2-3}{27}=-\frac{1}{27}.
     $$
     
   - The coordinates of the other turning point are $$\left(-\frac{1}{3}, -\frac{1}{27}\right)$$.

2. **Determining the value of $$a$$ from the $$x$$-axis intersections:**

- Find the $$x$$-intercepts by setting $$f(x)=0$$:  
     $$
     -2x^3-x^2=0 \quad \Longrightarrow \quad -x^2(2x+1)=0
     $$
     
   - Solve the factors:  
     $$
     -x^2=0 \quad \Longrightarrow \quad x=0, \qquad 2x+1=0 \quad \Longrightarrow \quad x=-\frac{1}{2}.
     $$
     
   - Since the intercepts are $$(0,0)$$ and $$(a,0)$$ with $$a=-\frac{1}{2}$$, we have:  
     $$
     a=-\frac{1}{2}.
     $$

3. **Sketch of the graph of $$f(x)$$:**

- **Intercepts:**  
     - $$x$$-intercepts at $$(0,0)$$ and $$\left(-\frac{1}{2},0\right)$$.  
     - $$y$$-intercept at $$f(0)=0$$ (i.e. $$(0,0)$$).
   
   - **Turning points:**  
     - $$(0,0)$$  
     - $$\left(-\frac{1}{3}, -\frac{1}{27}\right)$$
     
   - Plot these key points and sketch a smooth cubic curve for $$f(x)=-2x^3-x^2$$ ensuring that the curve touches the $$x$$-axis at the intercepts and has the noted turning behavior.

**5.1 Graph of the ellipse $$9x^2+4y^2=36$$:** - **Intercepts:** - $$x$$-intercepts at $$(2,0)$$ and $$(-2,0)$$. - $$y$$-intercepts at $$(0,3)$$ and $$(0,-3)$$. - **Graph:** An ellipse centered at $$(0,0)$$ with horizontal semi-axis of length 2 and vertical semi-axis of length 3. **5.2 Derivative of $$y=\frac{2}{x^{2}}+\sqrt{x}+\frac{1}{x}\pi$$:** $$ \frac{dy}{dx} = -\frac{4}{x^3} + \frac{1}{2\sqrt{x}} - \frac{\pi}{x^2} $$ **D.3.3 Analysis of $$f(x)=-2x^3-x^2$$:** - **Turning Points:** - $$(0,0)$$ - $$\left(-\frac{1}{3}, -\frac{1}{27}\right)$$ - **Value of $$a$$:** $$ a = -\frac{1}{2} $$ - **Graph:** A cubic curve intersecting the $$x$$-axis at $$(0,0)$$ and $$\left(-\frac{1}{2},0\right)$$, with turning points at $$(0,0)$$ and $$\left(-\frac{1}{3}, -\frac{1}{27}\right)$$.

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