The diagram shows the graph of $y = f(x)$. (a) Draw separate one-third page sketches of the graphs of the following: (i) $y = f(x + 3)$ (ii) $y = |f(x)|$ (iii) $y = \sqrt{f(x)}$ (iv) $y = f(|x|)$ (b) Sketch the graph of $y = x + \frac{8x}{x^2 - 9}$, clearly indicating any asymptotes and any points where the graph meets the axes. (c) Find the equation of the normal to the curve $x^3 - 4xy + y^3 = 1$ at $(2, 1)$. (d) The diagram shows the forces acting on a point $P$ which is moving on a frictionless banked circular track. The point $P$ has mass $m$ and is moving in a horizontal circle of radius $r$ with uniform speed $v$. The track is inclined at an angle $\theta$ to the horizontal. The point experiences a normal reaction force $N$ from the track and a vertical force of magnitude $mg$ due to gravity, so that the net force on the particle is a force of magnitude $\frac{mp^2}{r}$ directed towards the centre of the horizontal circle. By resolving $N$ in the horizontal and vertical directions, show that $N = m \sqrt{g^2 + \frac{v^4}{r^2}}$.

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The diagram shows the graph of $y = f(x)$ - HSC - SSCE Mathematics Extension 2 - Question 3 - 2005 - Paper 1

Question 3

The diagram shows the graph of $y = f(x)$. (a) Draw separate one-third page sketches of the graphs of the following: (i) $y = f(x + 3)$ (ii) $y = |f(x)|$ (iii) $... show full transcript

Worked Solution & Example Answer:The diagram shows the graph of $y = f(x)$ - HSC - SSCE Mathematics Extension 2 - Question 3 - 2005 - Paper 1

Step 1

Draw separate one-third page sketches of the graphs of the following: (i) $y = f(x + 3)$

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Answer

To sketch the graph of $y = f(x + 3)$ , you will translate the original graph of $y = f(x)$ to the left by 3 units. This shift affects only the horizontal position of the graph, maintaining its shape and vertical position, resulting in a new graph that starts and ends at the same respective vertical points but is horizontally displaced.

Step 2

Draw separate one-third page sketches of the graphs of the following: (ii) $y = |f(x)|$

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Answer

For the graph of $y = |f(x)|$ , reflect any parts of the graph of $f(x)$ that are below the x-axis. The resulting graph will be non-negative, illustrating the distances of $f(x)$ above the x-axis, while preserving the sections of the graph that are above it. Identify points of intersection and ensure continuity where relevant.

Step 3

Draw separate one-third page sketches of the graphs of the following: (iii) $y = \sqrt{f(x)}$

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Answer

To sketch $y = \sqrt{f(x)}$ , first determine where $f(x)$ is non-negative, as the square root function is only defined for non-negative inputs. Where $f(x) > 0$ , plot the square root values, preserving the shape but compressing the vertical distance of the graph. Where $f(x) < 0$ , there will be no graph in those regions.

Step 4

Draw separate one-third page sketches of the graphs of the following: (iv) $y = f(|x|)$

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Answer

The graph of $y = f(|x|)$ entails reflecting the portion of the graph of $f(x)$ that lies to the right of the y-axis across the y-axis to the left side. This graph will reflect symmetry about the y-axis for any parts of $f(x)$ that are defined on positive $x$ .

Step 5

Sketch the graph of $y = x + \frac{8x}{x^2 - 9}$, clearly indicating any asymptotes and points where the graph meets the axes.

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Answer

To sketch this graph, first find the vertical asymptotes by setting the denominator $x^2 - 9 = 0$ , which gives $x = 3$ and $x = -3$ . Identify horizontal behavior as $x \to \pm \infty$ for the end behavior of the graph. Evaluate $y$ -intercepts by checking $x = 0$ , resulting in $(0, 0)$ . Mark the asymptotes and intercepts on the graph to illustrate the overall behavior.

Step 6

Find the equation of the normal to the curve $x^3 - 4xy + y^3 = 1$ at $(2, 1)$.

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Answer

To find the normal, implicitly differentiate the equation to find $rac{dy}{dx}$ . At the point $(2, 1)$ , substitute these values to calculate the slope of the tangent line. The normal line's slope will be the negative reciprocal of this slope. Finally, using the point-slope form and the given point, write the equation of the normal line.

Step 7

By resolving $N$ in the horizontal and vertical directions, show that $N = m \sqrt{g^2 + \frac{v^4}{r^2}}$.

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Answer

To solve for $N$ , resolve the forces acting at point $P$ . The vertical forces yield the equation $mg = N \sin(\theta)$ and the horizontal forces yield $N \cos(\theta) = \frac{mv^2}{r}$ . Substituting to eliminate $N$ , manipulate the resulting equations to arrive at the required expression $N = m \sqrt{g^2 + \frac{v^4}{r^2}}$ .

The diagram shows the graph of $y = f(x)$ - HSC - SSCE Mathematics Extension 2 - Question 3 - 2005 - Paper 1

Worked Solution & Example Answer:The diagram shows the graph of $y = f(x)$ - HSC - SSCE Mathematics Extension 2 - Question 3 - 2005 - Paper 1

Draw separate one-third page sketches of the graphs of the following: (i) $y = f(x + 3)$

Draw separate one-third page sketches of the graphs of the following: (ii) $y = |f(x)|$

Draw separate one-third page sketches of the graphs of the following: (iii) $y = \sqrt{f(x)}$

Draw separate one-third page sketches of the graphs of the following: (iv) $y = f(|x|)$

Sketch the graph of $y = x + \frac{8x}{x^2 - 9}$, clearly indicating any asymptotes and points where the graph meets the axes.

Find the equation of the normal to the curve $x^3 - 4xy + y^3 = 1$ at $(2, 1)$.

By resolving $N$ in the horizontal and vertical directions, show that $N = m \sqrt{g^2 + \frac{v^4}{r^2}}$.

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