A function \( f \) is defined by \( f(x)=\frac{x^{2}+6 x+12}{x+2}, x \neq-2 \). (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 \( y=3-2|f(x)| \)

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,