QUESTION 5 Given: \( f(x)=\frac{4}{x-3}+2 \) and \( g(x)=x+2 \) \( 5.1 \quad \) Write down the equations of the asymptotes of \( f \). \( 5.2 \quad \) Determine the \( x \)-intercept of \( f \). 5.3 \( \begin{array}{ll}\text { Determine the } y \text {-intercept of } f . \\ 5.4 & \begin{array}{l}\text { Sketch the graphs of } f \text { and } g \text { on the same system of axes. Show clearly ALL the } \\ \text { intercepts with the axes and any asymptotes. } \\ 5.5\end{array} \quad \text { Calculate the } x \text {-coordinates of the points of intersection of } f \text { and } g \text {. } \\ 5.6 & \text { If } x<3, \text { determine the values of } x \text { for which } \frac{4}{x-3}+2

Ask by Parry Martinez. in South Africa

Feb 03,2025

Let's solve the problem step by step. ### 5.1: Write down the equations of the asymptotes of \( f \). The function \( f(x) = \frac{4}{x-3} + 2 \) has two types of asymptotes: vertical and horizontal. 1. **Vertical Asymptote**: This occurs where the denominator is zero. \[ x - 3 = 0 \implies x = 3 \] So, the vertical asymptote is \( x = 3 \). 2. **Horizontal Asymptote**: As \( x \) approaches infinity, the term \( \frac{4}{x-3} \) approaches 0. Therefore, the horizontal asymptote is: \[ y = 2 \] **Asymptotes**: - Vertical: \( x = 3 \) - Horizontal: \( y = 2 \) ### 5.2: Determine the \( x \)-intercept of \( f \). To find the \( x \)-intercept, we set \( f(x) = 0 \): \[ \frac{4}{x-3} + 2 = 0 \] Subtracting 2 from both sides: \[ \frac{4}{x-3} = -2 \] Multiplying both sides by \( x - 3 \) (noting that \( x \neq 3 \)): \[ 4 = -2(x - 3) \] Expanding: \[ 4 = -2x + 6 \] Rearranging gives: \[ 2x = 6 - 4 \implies 2x = 2 \implies x = 1 \] Thus, the \( x \)-intercept is \( (1, 0) \). ### 5.3: Determine the \( y \)-intercept of \( f \). To find the \( y \)-intercept, we set \( x = 0 \): \[ f(0) = \frac{4}{0-3} + 2 = \frac{4}{-3} + 2 = -\frac{4}{3} + 2 = -\frac{4}{3} + \frac{6}{3} = \frac{2}{3} \] Thus, the \( y \)-intercept is \( \left(0, \frac{2}{3}\right) \). ### 5.4: Sketch the graphs of \( f \) and \( g \). To sketch the graphs, we need to plot the asymptotes, intercepts, and the function \( g(x) = x + 2 \). - **Asymptotes**: - Vertical: \( x = 3 \) - Horizontal: \( y = 2 \) - **Intercepts of \( f \)**: - \( x \)-intercept: \( (1, 0) \) - \( y \)-intercept: \( \left(0, \frac{2}{3}\right) \) - **Intercepts of \( g \)**: - \( g(0) = 2 \) gives the \( y \)-intercept \( (0, 2) \). - \( g(x) = 0 \) gives the \( x \)-intercept \( (-2, 0) \). ### 5.5: Calculate the \( x \)-coordinates of the points of intersection of \( f \) and \( g \). To find the points of intersection, we set \( f(x) = g(x) \): \[ \frac{4}{x-3} + 2 = x + 2 \] Subtracting 2 from both sides: \[ \frac{4}{x-3} = x \] Multiplying both sides by \( x - 3 \): \[ 4 = x(x - 3) \] This simplifies to: \[ x^2 - 3x - 4 = 0 \] Now, we will solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \] Calculating the discriminant: \[ = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm \sqrt{25}}{2} = \frac{3 \pm 5}{2} \] This gives us two solutions: \[ x = \frac{8}{2} = 4 \quad \text{and} \quad x = \frac{-2}{2} = -1 \] Thus, the \( x \)-coordinates of the points of intersection are \( x = 4 \) and \( x = -1 \). ### 5.6: If \( x < 3 \), determine the values of \( x \) for which \( \frac{4}{x-3} + 2 < x + 2 \). We already have: \[ \frac{4}{x-3} < x \] Multiplying both sides by \( x - 3 \) (noting that \( x < 3 \) makes \( x - 3 < 0 \), so we reverse the inequality): \[ 4 > x(x - 3) \] This simplifies to: \[ x^2 - 3x - 4 < 0 \] Factoring gives: \[ (x - 4)(x + 1) < 0 \] The critical points are \( x = -1 \) and \( x = 4 \). Testing intervals: - For \( x < -1 \): both factors are negative, product is positive. - For \( -1 < x < 4 \): one factor is negative, product is negative. - For \( x > 4 \): both factors are positive, product is positive. Thus, the solution is: \[ -1 < x < 4 \] Since we are considering \( x < 3 \), the final solution is: \[ -1 < x < 3 \] ### 5.7: The line \( y = x - 1 \) cuts \( f \) at \( P(1, 0) \) and \( Q \). Write down the coordinates of \( Q \). To find \( Q \), we set \( f(x) = x - 1 \): \[ \frac{4}{x-3} + 2 = x - 1 \] Subtracting 2 from both sides: \[ \frac{4}{x-3} = x - 3 \] Multiplying both sides by \( x - 3 \): \[ 4 = (x - 3)(x - 3