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

The function \( f(x) = \frac{4}{x-3} + 2 \) has a vertical asymptote where the denominator is zero, which occurs at \( x = 3 \). This means that as \( x \) approaches 3, \( f(x) \) will increase or decrease without bound. Additionally, there is a horizontal asymptote at \( y = 2 \) because as \( x \) approaches positive or negative infinity, the term \( \frac{4}{x-3} \) approaches 0. To find the \( x \)-intercept of \( f \), set \( f(x) = 0 \): \[ 0 = \frac{4}{x-3} + 2 \] Solving gives \[ \frac{4}{x-3} = -2 \implies 4 = -2(x - 3) \implies x - 3 = -2 \implies x = 1 \] So, the \( x \)-intercept is at \( (1, 0) \). For the \( y \)-intercept, set \( x = 0 \): \[ f(0) = \frac{4}{0-3} + 2 = -\frac{4}{3} + 2 = \frac{2}{3} \] Thus, the \( y \)-intercept is \( (0, \frac{2}{3}) \). To sketch the graphs, plot the asymptotes at \( x = 3 \) (dashed vertical line) and \( y = 2 \) (dashed horizontal line). The graph of \( f \) approaches these lines, and the intercepts show where it crosses the axes. The line \( g(x) = x + 2 \) is a straight line with a \( y \)-intercept of 2 and a slope of 1, coming in from below the horizontal asymptote at \( y = 2 \). To find \( x \)-coordinates of the intersection of \( f(x) \) and \( g(x) \), set the equations equal: \[ \frac{4}{x-3} + 2 = x + 2 \] This simplifies to \( \frac{4}{x-3} = x \); cross-multiplying yields \( 4 = x(x-3) \). Rearranging gives: \[ x^2 - 3x - 4 = 0 \] Factoring leads to \( (x-4)(x+1) = 0 \), so \( x = 4 \) or \( x = -1 \). Next, for \( \frac{4}{x-3} + 2 < x + 2 \): Subtracting 2 gives \( \frac{4}{x-3} < x \), resulting in \( 4 < x(x-3) \), or \( 0 < x^2 - 3x - 4 \). Solving the quadratic \( x^2 - 3x - 4 = 0 \) leads again to \( (x-4)(x+1) = 0 \). Thus, the intervals where the inequality holds are \( x < -1 \) or \( x > 4 \) (keeping \( x < 3 \)). So, the valid solution is \( x < -1 \). For the point \( Q \) where the line \( y = x - 1 \) cuts \( f \), substitute \( x - 1 \) into \( f \). Setting equations equal gives: \[ \frac{4}{x-3} + 2 = x - 1 \] Solving this yields \( x = 5 \) for \( P \). Substituting into the line equation \( y \) gives \( Q(5, 4) \).