Given: \( f(x)=\frac{4}{x-3}+2 \) and \( g(x)=x+2 \) 5.1 Write down the equations of the asymptotes of \( f \). 5.2 Determine the \( x \)-intercept of \( f \). 5.3 Determine the \( y \)-intercept of \( f \). 5.4 Sketch the graphs of \( f \) and \( g \) on the same system of axes. Show clearly ALL the intercepts with the axes and any asymptotes. 5.5 Calculate the \( x \)-coordinates of the points of intersection of \( f \) and \( g \). 5.6 If \( x<3 \), determine the values of \( x \) for which \( \frac{4}{x-3}+2

Ask by Potter Tyler. in South Africa

Feb 03,2025

Let's tackle each part of the problem step-by-step. --- ### **Given:** - \( f(x) = \frac{4}{x - 3} + 2 \) - \( g(x) = x + 2 \) --- ### **5.1. Equations of the Asymptotes of \( f \)** **Vertical Asymptote:** The vertical asymptote occurs where the denominator of the rational function is zero. \[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \] **Horizontal Asymptote:** For rational functions of the form \( \frac{A}{x - h} + k \), as \( x \) approaches infinity or negative infinity, the \(\frac{A}{x - h}\) term approaches 0. Therefore, the horizontal asymptote is the constant term added to it. \[ y = 2 \] **Summary:** - **Vertical Asymptote:** \( x = 3 \) - **Horizontal Asymptote:** \( y = 2 \) --- ### **5.2. \( x \)-Intercept of \( f \)** The \( x \)-intercept is found by setting \( f(x) = 0 \): \[ \frac{4}{x - 3} + 2 = 0 \] \[ \frac{4}{x - 3} = -2 \] \[ 4 = -2(x - 3) \] \[ 4 = -2x + 6 \] \[ -2 = -2x \] \[ x = 1 \] **\( x \)-Intercept:** \( (1, 0) \) --- ### **5.3. \( y \)-Intercept of \( f \)** The \( y \)-intercept is found by evaluating \( f(0) \): \[ f(0) = \frac{4}{0 - 3} + 2 = \frac{4}{-3} + 2 = -\frac{4}{3} + 2 = -\frac{4}{3} + \frac{6}{3} = \frac{2}{3} \] **\( y \)-Intercept:** \( \left(0, \frac{2}{3}\right) \) --- ### **5.4. Sketching the Graphs of \( f \) and \( g \)** **For \( f(x) = \frac{4}{x - 3} + 2 \):** - **Vertical Asymptote:** \( x = 3 \) - **Horizontal Asymptote:** \( y = 2 \) - **\( x \)-Intercept:** \( (1, 0) \) - **\( y \)-Intercept:** \( \left(0, \frac{2}{3}\right) \) **For \( g(x) = x + 2 \):** - **Slope:** 1 - **\( y \)-Intercept:** \( (0, 2) \) - **\( x \)-Intercept:** Set \( g(x) = 0 \): \[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \] So, \( (-2, 0) \) **Graph Description:** - **\( f(x) \)** will have a hyperbola shape approaching the asymptotes \( x = 3 \) and \( y = 2 \), passing through \( (1, 0) \) and \( \left(0, \frac{2}{3}\right) \). - **\( g(x) \)** is a straight line crossing the \( y \)-axis at \( (0, 2) \) and the \( x \)-axis at \( (-2, 0) \). *Since this is a text medium, please plot the functions on graph paper or using graphing software to visualize their intersections, intercepts, and asymptotes clearly.* --- ### **5.5. \( x \)-Coordinates of Points of Intersection of \( f \) and \( g \)** To find the points where \( f(x) = g(x) \): \[ \frac{4}{x - 3} + 2 = x + 2 \] Subtract 2 from both sides: \[ \frac{4}{x - 3} = x \] Multiply both sides by \( (x - 3) \) (keeping in mind \( x \neq 3 \)): \[ 4 = x(x - 3) \] \[ 4 = x^2 - 3x \] Bring all terms to one side: \[ x^2 - 3x - 4 = 0 \] Solve the quadratic equation: \[ x = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm \sqrt{25}}{2} = \frac{3 \pm 5}{2} \] **Solutions:** - \( x = \frac{3 + 5}{2} = 4 \) - \( x = \frac{3 - 5}{2} = -1 \) **Points of Intersection:** - \( (4, 6) \) since \( g(4) = 4 + 2 = 6 \) - \( (-1, 1) \) since \( g(-1) = -1 + 2 = 1 \) --- ### **5.6. Values of \( x \) for Which \( \frac{4}{x - 3} + 2 < x + 2 \) When \( x < 3 \)** We need to solve the inequality: \[ \frac{4}{x - 3} + 2 < x + 2 \] Subtract 2 from both sides: \[ \frac{4}{x - 3} < x \] **Important:** Since \( x < 3 \), \( x - 3 < 0 \). When multiplying or dividing by a negative number, the inequality sign reverses. Multiply both sides by \( (x - 3) \) (which is negative): \[ 4 > x(x - 3) \] \[ 4 > x^2 - 3x \] Bring all terms to one side: \[ x^2 - 3x - 4 < 0 \] **Solve the Quadratic Inequality:** First, find the roots of \( x^2 - 3x - 4 = 0 \): \[ x = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2} \] \[ x = 4 \quad \text{and} \quad x = -1 \] The quadratic \( x^2 - 3x - 4 \) is less than 0 between its roots: \[ -1 < x < 4 \] **But** since we're given \( x < 3 \), we intersect this with \( -1 < x < 4 \): \[ -1 < x < 3 \] **Conclusion:** All real numbers \( x \) such that \( -1 < x < 3 \) satisfy the inequality \( \frac{4}{x - 3} + 2 < x + 2 \). ---