15 REFLECTIONSCONSIDER THE FIRST QUADRANT point (a, b), and let us reflect it about the y-axis. It is reflected to the second quadrant point (−a, b). If we reflect (a, b) about the x-axis, then it is reflected to the fourth quadrant point (a, −b). Finally, if we reflect (a, b) through the origin, then it is reflected to the third quadrant point (−a, −b). The distance from the origin to (a, b) is equal to the distance from the origin to (−a, −b). Example 1. Fig. 1 is the graph of the parabola f(x) = x² − 2x − 3 = (x + 1)(x − 3). The roots −1, 3 are the x-intercepts. Fig. 2 is its reflection about the x-axis. Every point that was above the x-axis gets reflected to below the x-axis. And every point below the x-axis gets reflected above the x-axis. Only the roots, −1 and 3, are invariant. Again, Fig. 1 is y = f(x). Its reflection about the x-axis is y = −f(x). Every y-value there is the negative of the original f(x). Fig. 3 is the reflection of Fig. 1 about the y-axis. Every point that was to the right of the origin gets reflected to the left. And every point that was on the left gets reflected to the right. Every x becomes −x. Only the y-intercept is invariant. The equation of the reflection of f(x) about the y-axis is y = f(−x). The argument x of f(x) is replaced by −x. See Problem 1c) below. If y = f(x), then y = f(−x) is its reflection about the y-axis, y = −f(x) is its reflection about the x-axis. Problem 1. Let f(x) = x² + x − 2. a) Sketch the graph of f(x). To see the answer, pass your mouse over the colored area. x² + x − 2 = (x + 2)(x − 1). The x-intercepts are at −2 and 1. b) Write the function −f(x), and sketch its graph. −f(x) = −(x² + x − 2) = −x² − x + 2. Its graph is the reflection of f(x) about the x-axis.
c) Write the function f(−x), and sketch its graph. Replace each x with −x. f(−x) = (−x)² − x − 2 = x² − x − 2 = (x − 2)(x + 1). Its graph is the reflection of f(x) about the y-axis.
Problem 2. Let f(x) = (x + 3)(x + 1)(x − 2). Sketch the graph of f(x), then sketch the graphs of f(−x) and −f(x).
The graph on the left is f(x). The roots -- the x-intercepts -- are −3, −1, 2. The middle graph is f(−x), which is its reflection about the y-axis. The graph on the right is −f(x), which is its reflection about the x-axis. Problem 3. Let f(x) = x² − 4. Sketch the graph of f(x), then sketch the graph of f(−x). x² − 4 = (x + 2)(x − 2). Here, the graph of f(−x) -- its reflection about the y-axis -- is equal to the graph of f(x). Problem 4. Let f(x) = x³. Sketch the graph of f(x), then sketch the graphs of f(−x) and −f(x).
The graph on the left is f(x). The graph on the right is f(−x), which is its reflection about the y-axis . But (−x )³ = −x³, so that Example 2. Sketch the graph of y = −x² + x + 6. Solution. It is best to consider a graph when the leading coefficient is positive. Therefore, let us call the given function −f(x):
f(x), then, is (x + 2)(x − 3). Its x-intercepts are at −2 and 3. The graph we want is −f(x), which is the reflection of f(x) about the x-axis: Problem 5. Sketch the graph of y = −x² − 2x + 8.
Here is the graph: It is the reflection about the x-axis of f(x) = x² + 2x − 8 Problem 6. Sketch the graph of y = −x^{3} − 2x² + x + 2. [Hint: Call the function −f(x), then factor f(x) by grouping.]
The graph is the reflection about the x-axis of f(x) = (x + 2)(x + 1)(x − 1) Problem 7. Sketch the graph of y = −. This is the reflection of the square root function about the x-axis. (See Topic 5.) Problem 8. Sketch the graph of y = −|x|. This is the reflection of the absolute value function about the x-axis. (See Topic 5.) Please make a donation to keep TheMathPage online. Copyright © 2014 Lawrence Spector Questions or comments? E-mail: themathpage@nyc.rr.com |