Question

Sketch the graph of the first function by plotting points if necessary. Then use transformation(s) to obtain the graph of the second function. $$y=x^{2}, \quad y=\left|x^{2}-2 x-1\right|$$ 54\. $$y=\tan x, \quad y=\tan \left(x+\frac{\pi}{3}\ri

Short answer

Short Answer: For the first pair of functions, transform the graph of \(y=x^2\) by translating it 1 unit right, 1 unit downward, and reflecting negative parts above the x-axis to obtain the graph of \(y=|x^2-2x-1|\). For the second pair, just translate the graph of \(y=\tan x\) \(\frac{\pi}{3}\) units to the left to obtain the graph of \(y=\tan(x+\frac{\pi}{3})\).

Step-by-step solution

Sketch the graph of the first function, \(y=x^2\)

Recall that the graph of \(y=x^2\) is a parabola with its vertex at the origin (0,0) and opening upwards. To sketch the graph, plot some points of the function, for example, x = -2, -1, 0, 1, and 2, and connect the points with a smooth curve, resulting in a parabolic shape.

Determine the transformation needed to obtain the second function

We need to find the relationship between the first function \(y=x^2\) and the second function \(y=|x^2-2x-1|\). The transformation applied here can be broken down as follows: 1. Replace \(y = x^2\) with \(y = x^2 - 2x -1\). This involves a horizontal translation to the right by 1 unit and a vertical translation downwards by 1 unit. 2. Take the absolute value of the entire expression, which is denoted by \(|x^2 - 2x - 1|\). This transformation reflects any negative part of the graph above the x-axis.

Apply the transformations to the graph

Following the transformations determined in Step 2: 1. Perform a horizontal translation of the parabola 1 unit to the right and a vertical translation 1 unit downward. This results in the vertex of the new parabola being at (1,-1). 2. Reflect the negative parts of the new parabola above the x-axis. This will create a V-shaped graph. #Pair 2: Involving Trigonometric Functions#

Sketch the graph of the first function, \(y=\tan x\)

Recall that the graph of \(y=\tan x\) has a periodic behavior with a period of \(\pi\). To sketch the graph, plot some points where x takes values in the interval \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) (excluding these exact values). Connect the points with a smooth curve, resulting in a series of branches, passing through the origin (0,0), and with asymptotes at \(\pm\frac{\pi}{2}\) and their multiples.

Determine the transformation needed to obtain the second function

We need to find the relationship between the first function \(y=\tan x\) and the second function \(y=\tan(x+\frac{\pi}{3})\). The transformation applied here can be broken down as follows: 1. Replace \(y = \tan x\) with \(y = \tan (x+\frac{\pi}{3})\). This involves a horizontal translation to the left by \(\frac{\pi}{3}\) units.

Apply the transformation to the graph

Apply the horizontal translation determined in Step 2 to the graph of \(y = \tan x\). Move the branches of the graph \(\frac{\pi}{3}\) units to the left. This will cause the asymptotes to shift to the left by \(\frac{\pi}{3}\) units, and the intercepts will move to the left by the same amount, resulting in the graph of \(y=\tan(x+\frac{\pi}{3})\).