Question
Consider the function defined by \( f(x) = \frac{x^2 – 14x + 24}{2x + 6} \), where \( x \in \mathbb{R}, \, x \neq -3 \).
(a) State the equation of the vertical asymptote on the graph of \( f \). [1]
(b) Find the coordinates of the points where the graph of \( f \) crosses the \( x \)-axis. [2]
(c) Find the value of \( a \) and the value of \( b \) for the oblique asymptote \( y = ax + b \). [4]
(d) Sketch the graph of \( f \) for \(-50 \leq x \leq 50\), showing clearly the asymptotes and any intersections with the axes. [4]
(e) Find the range of \( f \). [4]
(f) Solve the inequality \( f(x) > x \). [4]
▶️Answer/Explanation
(a) The vertical asymptote occurs where the denominator is zero: \[ 2x + 6 = 0 \implies x = -3 \] So, the equation of the vertical asymptote is: \[ x = -3 \]
(b) The graph crosses the \( x \)-axis when \( f(x) = 0 \): \[ x^2 – 14x + 24 = 0 \implies (x – 2)(x – 12) = 0 \] So, the coordinates are: \[ (2, 0) \, \text{and} \, (12, 0) \]
(c) To find the oblique asymptote, perform polynomial long division: \[ f(x) = \frac{x^2 – 14x + 24}{2x + 6} = \frac{1}{2}x – \frac{17}{2} + \frac{75}{2x + 6} \] As \( x \to \infty \), the term \( \frac{75}{2x + 6} \to 0 \), so the oblique asymptote is: \[ y = \frac{1}{2}x – \frac{17}{2} \] Thus, \( a = \frac{1}{2} \) and \( b = -\frac{17}{2} \).
(d) The graph of \( f(x) \) should show: – A vertical asymptote at \( x = -3 \). – An oblique asymptote at \( y = \frac{1}{2}x – \frac{17}{2} \). – \( x \)-intercepts at \( (2, 0) \) and \( (12, 0) \). – The graph should approach the asymptotes as \( x \to \pm \infty \).
(e) The range of \( f \) is all real numbers except where the function is undefined. From the graph, the range is: \[ y \leq -18.7 \, \text{or} \, y \geq -1.34 \]
(f) Solve \( f(x) > x \): \[ \frac{x^2 – 14x + 24}{2x + 6} > x \] Rearranging: \[ \frac{x^2 – 14x + 24 – x(2x + 6)}{2x + 6} > 0 \implies \frac{-x^2 – 20x + 24}{2x + 6} > 0 \] Solving the inequality: \[ x < -21.1 \, \text{or} \, -3 < x < 1.14 \]