For \( f(x) = \frac{4x + 2}{x – 2} \), \( x \neq 2 \), sketch the graph with intercepts and asymptotes.

Vertical Asymptote: Denominator is zero at \( x – 2 = 0 \), so \( x = 2 \). Numerator at \( x = 2 \): \( 4 \cdot 2 + 2 = 10 \neq 0 \). Thus, vertical asymptote at \( x = 2 \). As \( x \to 2^- \), \( x – 2 \to 0^- \), \( y \to -\infty \). As \( x \to 2^+ \), \( x – 2 \to 0^+ \), \( y \to +\infty \).

Horizontal Asymptote: Degrees of numerator and denominator are equal, so the asymptote is the ratio of leading coefficients: \( \frac{4}{1} = 4 \). Thus, \( y = 4 \). Confirm: \( \lim_{x \to \pm \infty} \frac{4x + 2}{x – 2} = \lim_{x \to \pm \infty} \frac{4 + \frac{2}{x}}{1 – \frac{2}{x}} = \frac{4}{1} = 4 \).

x-Intercept: Set \( f(x) = 0 \): \( \frac{4x + 2}{x – 2} = 0 \implies 4x + 2 = 0 \implies 4x = -2 \implies x = -\frac{1}{2} \). Check: \( x – 2 = -\frac{1}{2} – 2 = -\frac{5}{2} \neq 0 \). So, x-intercept is \( \left(-\frac{1}{2}, 0\right) \).

y-Intercept: Set \( x = 0 \): \( f(0) = \frac{4 \cdot 0 + 2}{0 – 2} = \frac{2}{-2} = -1 \). So, y-intercept is \( (0, -1) \).

Behavior: Test points: At \( x = 1 \), \( f(1) = \frac{4 + 2}{1 – 2} = -6 \). At \( x = 3 \), \( f(3) = \frac{12 + 2}{3 – 2} = 14 \). Graph has two branches: left of \( x = 2 \), it’s below \( y = 4 \), crossing \( \left(-\frac{1}{2}, 0\right) \) and \( (0, -1) \), approaching \( -\infty \). Right of \( x = 2 \), it starts at \( +\infty \), descends toward \( y = 4 \).

Answer: Graph with vertical asymptote \( x = 2 \), horizontal asymptote \( y = 4 \), x-intercept \( \left(-\frac{1}{2}, 0\right) \), y-intercept \( (0, -1) \), showing correct shape and asymptotic behavior.

Find the range of \( f(x) = \frac{4x + 2}{x – 2} \). Solve for \( x \) in terms of \( y \):

\[ y = \frac{4x + 2}{x – 2} \]

\[ y (x – 2) = 4x + 2 \]

\[ yx – 2y = 4x + 2 \]

\[ yx – 4x = 2y + 2 \]

\[ x (y – 4) = 2y + 2 \]

\[ x = \frac{2y + 2}{y – 4}, \quad y \neq 4 \]

Denominator zero at \( y = 4 \), matching the horizontal asymptote. Check if \( x = 2 \) (undefined) produces a \( y \)-value: \( 2 = \frac{2y + 2}{y – 4} \implies 2y – 8 = 2y + 2 \implies -8 = 2 \), a contradiction. Thus, no \( y \) maps to \( x = 2 \). As \( x \) covers \( (-\infty, 2) \cup (2, \infty) \), \( y \) covers all reals except \( y = 4 \).

Answer: \( y \in \mathbb{R} \setminus \{4\} \)

For \( g(x) = x^2 + bx + c \), the axis of symmetry is \( x = 2 \), roots are \( -\frac{1}{2} \) and \( p \). Axis of symmetry: \( x = -\frac{b}{2a} \). With \( a = 1 \):

\[ -\frac{b}{2} = 2 \implies b = -4 \]

Sum of roots: \( -\frac{1}{2} + p = -\frac{b}{a} = -(-4) = 4 \).

\[ p – \frac{1}{2} = 4 \implies p = 4 + \frac{1}{2} = \frac{9}{2} \]

Since \( \frac{9}{2} \in \mathbb{Q} \), the result holds.

Answer: \( p = \frac{9}{2} \)

From (c), \( b = -4 \). Product of roots gives \( c \):

\[ \left(-\frac{1}{2}\right) \cdot \frac{9}{2} = \frac{c}{a} = c \quad (a = 1) \]

\[ -\frac{9}{4} = c \]

Verify: \( g(x) = x^2 – 4x – \frac{9}{4} \). Roots via quadratic formula:

\[ x = \frac{4 \pm \sqrt{16 – 4 \cdot 1 \cdot (-\frac{9}{4})}}{2} = \frac{4 \pm \sqrt{16 + 9}}{2} = \frac{4 \pm 5}{2} = \frac{9}{2}, -\frac{1}{2} \]

Answer: \( b = -4 \), \( c = -\frac{9}{4} \)

Vertex x-coordinate is \( x = 2 \). Compute \( g(2) \):

\[ g(2) = 2^2 – 4 \cdot 2 – \frac{9}{4} = 4 – 8 – \frac{9}{4} = -4 – \frac{9}{4} = -\frac{16}{4} – \frac{9}{4} = -\frac{25}{4} \]

Answer: \( y = -\frac{25}{4} \)

Solve \( f(x) = g(x) \):

\[ \frac{4x + 2}{x – 2} = x^2 – 4x – \frac{9}{4} \]

Multiply by \( x – 2 \):

\[ 4x + 2 = \left(x^2 – 4x – \frac{9}{4}\right) (x – 2) \]

Right side: \( x^3 – 2x^2 – 4x^2 + 8x – \frac{9}{4}x + \frac{18}{4} = x^3 – 6x^2 + 8x + \frac{9}{2} \).

Equation: \( 4x + 2 = x^3 – 6x^2 + 8x + \frac{9}{2} \).

\[ 0 = x^3 – 6x^2 + 8x + \frac{9}{2} – 4x – 2 = x^3 – 6x^2 + 4x + \frac{5}{2} \]

Solve the cubic \( x^3 – 6x^2 + 4x + \frac{5}{2} = 0 \). Use Rational Root Theorem: Test \( x = -\frac{1}{2} \):

\[ -\frac{1}{8} – 6 \cdot \frac{1}{4} + 4 \cdot (-\frac{1}{2}) + \frac{5}{2} = -\frac{1}{8} – \frac{12}{8} – \frac{16}{8} + \frac{20}{8} = -\frac{9}{8} \neq 0 \]

Graphically: \( g(x) \) is a parabola, vertex at \( (2, -\frac{25}{4}) \), roots at \( -\frac{1}{2} \), \( \frac{9}{2} \). \( f(x) \) has a vertical asymptote at \( x = 2 \). As \( x \to 2^- \), \( f(x) \to -\infty \), \( g(2) = -\frac{25}{4} \), so \( f(x) – g(x) \to -\infty \). As \( x \to 2^+ \), \( f(x) \to +\infty \). Between \( -\frac{1}{2} \) and \( \frac{9}{2} \), \( g(x) < 0 \), while \( f(x) \) transitions through intercepts. Graph shows three intersections.

Answer: 3 solutions