Answer: (a) \(x=-\frac{3}{2}-\frac{\sqrt{5}}{2}\), \(x=-\frac{3}{2}+\frac{\sqrt{5}}{2}\), and \(y=0\).

(b) No stationary points.

(c) Intersections: \((0,2)\) and \((-2,0)\). Asymptotes: \(x=-\frac{3}{2}\pm\frac{\sqrt{5}}{2}\), \(y=0\).

(d) Reflect all parts of \(C\) below the \(x\)-axis above the \(x\)-axis.

(e) \(-\frac{7}{4}-\frac{\sqrt{17}}{4}\lt x\lt-\frac{3}{2}-\frac{\sqrt{5}}{2}\), \(-\frac{3}{2}-\frac{\sqrt{5}}{2}\lt x\lt-\frac{5}{2}\), \(-\frac{7}{4}+\frac{\sqrt{17}}{4}\lt x\lt-\frac{3}{2}+\frac{\sqrt{5}}{2}\), or \(-\frac{3}{2}+\frac{\sqrt{5}}{2}\lt x\lt0\).

(a)

Vertical asymptotes occur where the denominator is zero:

\(x^2+3x+1=0.\)

Thus

\[ x=\frac{-3\pm\sqrt{9-4}}{2} =-\frac{3}{2}\pm\frac{\sqrt{5}}{2}. \]

Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is

\(y=0.\)

(b)

Differentiate using the quotient rule:

\[ \frac{dy}{dx} = \frac{(x^2+3x+1)(1)-(x+2)(2x+3)}{(x^2+3x+1)^2}. \]

Stationary points would require the numerator to be zero:

\[ x^2+3x+1-(x+2)(2x+3)=0. \]

This simplifies to

\[ -x^2-4x-5=0, \]

or

\[ x^2+4x+5=0. \]

The discriminant is

\[ 4^2-4(1)(5)=16-20=-4\lt0, \]

so there are no real roots and hence no stationary points.

(c)

The \(y\)-intercept is found by setting \(x=0\):

\[ y=\frac{2}{1}=2, \]

so the curve passes through \((0,2)\).

The \(x\)-intercept is found by setting the numerator to zero:

\[ x+2=0\Rightarrow x=-2, \]

so the curve passes through \((-2,0)\).

The sketch should show vertical asymptotes at \(x=-\frac{3}{2}\pm\frac{\sqrt{5}}{2}\), horizontal asymptote \(y=0\), and the correct branch positions.

(d)

The graph of \(y=\left|f(x)\right|\) is obtained from \(y=f(x)\) by leaving all parts above the \(x\)-axis unchanged and reflecting all parts below the \(x\)-axis in the \(x\)-axis.

For this curve, the asymptotes stay the same, and the point \((-2,0)\) remains on the graph, forming the correct sharp point after reflection.

(e)

Solve the boundary equation:

\[ \left|\frac{x+2}{x^2+3x+1}\right|=2. \]

This gives

\[ \frac{x+2}{x^2+3x+1}=2 \quad\text{or}\quad \frac{x+2}{x^2+3x+1}=-2. \]

For \(\frac{x+2}{x^2+3x+1}=2\):

\[ x+2=2x^2+6x+2 \Rightarrow -2x^2-5x=0. \]

So

\[ x=0 \quad\text{or}\quad x=-\frac{5}{2}. \]

For \(\frac{x+2}{x^2+3x+1}=-2\):

\[ x+2=-2x^2-6x-2 \Rightarrow 2x^2+7x+4=0. \]

So

\[ x=-\frac{7}{4}\pm\frac{\sqrt{17}}{4}. \]

The expression is undefined at

\[ x=-\frac{3}{2}\pm\frac{\sqrt{5}}{2}. \]

Using the sketch or sign intervals gives

\[ -\frac{7}{4}-\frac{\sqrt{17}}{4}\lt x\lt-\frac{3}{2}-\frac{\sqrt{5}}{2}, \]

\[ -\frac{3}{2}-\frac{\sqrt{5}}{2}\lt x\lt-\frac{5}{2}, \]

\[ -\frac{7}{4}+\frac{\sqrt{17}}{4}\lt x\lt-\frac{3}{2}+\frac{\sqrt{5}}{2}, \]

\[ -\frac{3}{2}+\frac{\sqrt{5}}{2}\lt x\lt0. \]

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