Function Transformations, Inverses, and Graphing

Question 1 Show that f(x) = x - 6 and g(x) = x + 6 are inverses of each other. Answer: Given: $ f(x) = x - 6 $, $ g(x) = x + 6 $ For inversion: $ f(x) = x - 6 $ $ x = f(x) - 6 $ $ x + 6 = f(x) $ So, the inverse of $ f(x) = g(x) $ Now check $ g(f(x)) = x $: $ g(f(x)) = g(x - 6) $ $ = (x - 6) + 6 $ $ = x $ Now check $ f(g(x)) = x $: $ f(g(x)) = f(x + 6) $ $ = (x + 6) - 6 $ $ = x $ Since $ g(f(x)) = x $ and $ f(g(x)) = x $, $ f(x) $ and $ g(x) $ are inverses of each other. Question 2 The function is related to one of the parent functions described in an earlier section. $ g(x) = \frac{1}{6} \sqrt{x} $ a) Identify the parent function f. b) Describe the sequence of transformations from f to g. c) Use function notation to write g in terms of f. Answer: Step 1: Given: $ g(x) = \frac{1}{6} \sqrt{x} $ Step 2: a) The function belongs to the family of square root functions. Parent function is given by: $ f(x) = \sqrt{x} $ b) Sequence of transformations from f to g: Horizontal stretch by a factor of $ \frac{1}{6} $ c) Writing $ g(x) $ in terms of $ f(x) $: $ g(x) = \frac{1}{6} f(x) $ Question 3 Graph the function by hand, not by plotting points, but by starting with the graph of one of the standard functions and then applying the appropriate transformations. $ y = 1 - 2x + x^2 $ Answer: Step 1: Given: $ y = 1 - 2x + x^2 $ To sketch: The graph of $ y = 1 - 2x + x^2 $ using transformations. Step 2: We need to identify the parent function of $ y = 1 - 2x + x^2 $ $ y = 1 - 2x + x^2 $ Rewrite: $ y = (x - 1)^2 $ Here the parent function is $ y = x^2 $. Using this, we will get the translation graph $ y = (x - 1)^2 $ Since $ y = (x - 1)^2 $ is a parabola with vertex (1, 0). The graph of $ y = x^2 $ with vertex (0, 0) is given by: [Graph of $ y = x^2 $] The graph of $ y = (x - 1)^2 $ is obtained by shifting the graph $ y = x^2 $ by 1 unit in the right direction. Thus, the graph of $ y = 1 - 2x + x^2 $ is given by: [Graph of $ y = (x - 1)^2 $] Question 4 Write an equation for the graphed function by using transformations of the graphs of one of the toolkit functions. Answer: Step 1: Parent function of the given graph is $ y = |x| $ The transformations are made in this function to obtain the given graph. First, we get $ y = -|x| $ Then shift the graph downwards by 3 units. Step 2: The function is $ y = -|x - 2| - 3 $ Check the value now using the graph: In place of our function: $ x = 5 $ $ y = -|5 - 2| - 3 $ $ y = -|3| - 3 $ $ y = -3 - 3 $ $ y = -6 $ Similarly, all other points are satisfied. Therefore, the function corresponding to given graph is $ y = -|x - 2| - 3 $. Question 5 Standard transformations can be used to help graph rational functions of the form $ f(x) = \frac{A}{Bx - C} + D $. Explain how the parameters A, B, C, and D relate to the graph of the rational function. Answer: Step 1: Conclude function $ f(x) $: given: has a vertical asymptote at $ \frac{C}{B} $. Has a horizontal asymptote at $ y = D $. Step 2: Additionally, we may conclude that: if $ \frac{A}{B} \neq 0 $, the function will have a $ y $-intercept at $ f(0) = \frac{A}{-C} + D $. We may conclude the $ x $-intercept as: $ \frac{A}{Bx - C} + D = 0 $ $ \frac{A}{Bx - C} = -D $ $ A = -D (Bx - C) $ $ A = -D Bx + D C $ $ D Bx = D C - A $ $ x = \frac{D C - A}{D B} $ The function will have a domain of $ (-\infty, \frac{C}{B}) \cup (\frac{C}{B}, \infty) $ if $ A \neq 0 $. The function will decrease continuously $ \frac{A}{B} < 0 $, the function will increase in its domain. The end behavior of the function $ y \rightarrow D $ as $ x \rightarrow \pm \infty $. Finally, the graph of $ f(x) $ is obtained from $ y = \frac{1}{x} $ by the following transformation: $ \frac{1}{x} \rightarrow \frac{1}{Bx} $: translate $ \frac{C}{B} $ units to the right. $ \frac{1}{Bx} \rightarrow \frac{A}{Bx} $: vertical stretch by a factor of $ \frac{A}{B} $. $ \frac{A}{Bx} \rightarrow \frac{A}{Bx - C} + D $: translate $ D $ units up. Simplifying the last expression will result in the formula for $ f(x) $ as given.