A function $$ f $$ is defined by $$ f(x)=\frac{x^{2}+6 x+12}{x+2}, x
eq-2 $$. (a) Express $$ f(x) $$ in the form $$ a x+b+\frac{b}{x+2} $$ stating the values of $$ a $$ and $$ b $$. (b) Write down an equation for each of the two asymptotes. (c) Show that $$ f(x) $$ has two stationary points. Determine the coordinates and the nature of the stationary points. (d) Sketch the graph of $$ f $$. (e) State the range of values of $$ k $$ such that the equation $$ f(x)=k $$ has no (f) Determine the coordinates of the stationary points of the function $$ y=3-2|f(x)| $$

Question

A function $$ f $$ is defined by $$ f(x)=\frac{x^{2}+6 x+12}{x+2}, x 
eq-2 $$. (a) Express $$ f(x) $$ in the form $$ a x+b+\frac{b}{x+2} $$ stating the values of $$ a $$ and $$ b $$. (b) Write down an equation for each of the two asymptotes. (c) Show that $$ f(x) $$ has two stationary points. Determine the coordinates and the nature of the stationary points. (d) Sketch the graph of $$ f $$. (e) State the range of values of $$ k $$ such that the equation $$ f(x)=k $$ has no (f) Determine the coordinates of the stationary points of the function $$ y=3-2|f(x)| $$

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Let's solve the problem step by step. ### Part (a) We need to express $$ f(x) $$ in the form $$ ax + b + \frac{c}{x+2} $$. 1. Start with the function: $$ f(x) = \frac{x^2 + 6x + 12}{x + 2} $$ 2. Perform polynomial long division to divide $$ x^2 + 6x + 12 $$ by $$ x + 2 $$. - Divide the leading term: $$ \frac{x^2}{x} = x $$. - Multiply $$ x $$ by $$ x + 2 $$: $$ x^2 + 2x $$. - Subtract: $$ (x^2 + 6x + 12) - (x^2 + 2x) = 4x + 12 $$ - Now divide $$ 4x + 12 $$ by $$ x + 2 $$: - Divide the leading term: $$ \frac{4x}{x} = 4 $$. - Multiply $$ 4 $$ by $$ x + 2 $$: $$ 4x + 8 $$. - Subtract: $$ (4x + 12) - (4x + 8) = 4 $$ 3. Thus, we have: $$ f(x) = x + 4 + \frac{4}{x + 2} $$ From this, we can identify $$ a = 1 $$, $$ b = 4 $$, and $$ c = 4 $$. ### Part (b) To find the asymptotes: 1. **Vertical Asymptote**: Occurs where the denominator is zero: $$ x + 2 = 0 \implies x = -2 $$ 2. **Horizontal Asymptote**: As $$ x \to \infty $$ or $$ x \to -\infty $$, the function approaches the linear term: $$ y = x + 4 \quad \text{(as } x \to \infty \text{)} $$ Thus, the equations for the asymptotes are: - Vertical: $$ x = -2 $$ - Horizontal: $$ y = x + 4 $$ ### Part (c) To find stationary points, we need to find the derivative $$ f'(x) $$ and set it to zero. 1. Differentiate $$ f(x) $$: $$ f(x) = x + 4 + \frac{4}{x + 2} $$ Using the quotient rule for $$ \frac{4}{x + 2} $$: $$ f'(x) = 1 - \frac{4}{(x + 2)^2} $$ 2. Set the derivative to zero: $$ 1 - \frac{4}{(x + 2)^2} = 0 $$ $$ \frac{4}{(x + 2)^2} = 1 \implies (x + 2)^2 = 4 \implies x + 2 = \pm 2 $$ This gives: $$ x + 2 = 2 \implies x = 0 \quad \text{and} \quad x + 2 = -2 \implies x = -4 $$ 3. Now, find the coordinates of the stationary points: - For $$ x = 0 $$: $$ f(0) = 0 + 4 + \frac{4}{2} = 4 + 2 = 6 $$ So, one stationary point is $$ (0, 6) $$. - For $$ x = -4 $$: $$ f(-4) = -4 + 4 + \frac{4}{-2} = 0 - 2 = -2 $$ So, the other stationary point is $$ (-4, -2) $$. 4. To determine the nature of the stationary points, we can check the second derivative or the sign of $$ f'(x) $$ around these points. ### Part (d) To sketch the graph of $$ f(x) $$: - Plot the asymptotes: $$ x = -2 $$ and $$ y = x + 4 $$. - Mark the stationary points: $$ (0, 6) $$ and $$ (-4, -2) $$. - The function approaches the asymptotes as $$ x \to -2 $$ and as $$ x \to \infty $$. ### Part (e) To find the range of values of $$ k $$ such that $$ f(x) = k $$ has no solutions, we need to analyze the range of $$ f(x) $$. 1. The function has a vertical asymptote at $$ x = -2 $$ and two stationary points. 2. The maximum value occurs at $$ (0, 6) $$ and the minimum at $$ (-4, -2) $$. 3. Therefore, the range of $$ f(x) $$ is $$ (-2, 6) $$. Thus, $$ k $$ must be outside this interval for $$ f(x) = k $$ to have no solutions: $$ k < -2 \quad \text{or} \quad k > 6 $$ ### Part (f) To find the stationary points of $$ y = 3 - 2|f(x)| $$: 1. The stationary points of $$ |f(x)| $$ occur at the same $$ x $$ values as $$ f(x) $$ since $$ f(x) $$ is continuous and differentiable except at $$ x = -2 $$. 2. The stationary points of $$ y = 3 - 2|f(x)| $$ will also occur at $$ x = 0 $$ and $$ x = -4 $$. 3. Calculate the coordinates: - For $$ x = 0 $$: $$ y = 3 - 2|f(0)| = 3 - 2 \cdot 6 = 3 - 12 = -9 $$ So, one stationary point is $$ (0, -9) $$. - For $$ x = -4 $$: $$ y = 3 - 2|f(-4)| = 3 - 2 \cdot 2 = 3 - 4 = -1 $$ So, the other stationary point is $$ (-4, -1) $$. ### Summary of Results - (a) $$ a = 1, b = 4 $$ - (b) Vertical: $$ x = -2 $$, Horizontal: $$ y = x + 4 $$ - (c) Stationary points: $$ (0, 6) $$ and $$ (-4, - **Part (a):** $$ a = 1 $$, $$ b = 4 $$ - **Part (b):** Vertical asymptote at $$ x = -2 $$, Horizontal asymptote at $$ y = x + 4 $$ - **Part (c):** Two stationary points at $$ (0, 6) $$ and $$ (-4, -2) $$ - **Part (d):** Graph with asymptotes and stationary points as described - **Part (e):** $$ k < -2 $$ or $$ k > 6 $$ - **Part (f):** Stationary points at $$ (0, -9) $$ and $$ (-4, -1) $$

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