This preview shows page 1. Sign up to view the full content.
Unformatted text preview: s the corresponding y-value. By visual inspection, you can see that the maximum yvalue is 4 and that the graph attains this value twice, at (–8,4) and (4,4). Similarly, the minimum value of y is –4 and the graph attains this value twice, at (–4,–4) and (8,–4); in both instances, the absolute value of y is 4. Thus, the absolute value of y is at its maximum at four different x-values. (E) The figure shows the graph of y = 2. For any real number x, f(x) = 2. Thus, regardless of what number is added to or subtracted from x, the result is still a number whose function is 2 (y = 2). (D) To determine the features of the x−2 transformed line, substitute 2 for x in the function:
f x −2 = x −2 −2 = x −2−2 = x −4 2 2 2. 3. (B) In the general equation y = mx + b, the slope (m) is given as 3. To determine b, substitute –3 for x and 3 for y, then solve for b: 3 = 3(–3) + b; 12 = b. (D) Line P slopes upward from left to right at an angle less than 45°. Thus, the line’s slope (m in the equation y = mx + b) is a positive fraction less than 1. Also, line P crosses the y-axis at a positive y-value (above the x-axis). Thus, the line’s y-intercept (b in the equation y = mx + b) is positive. Only choice (D) provides an equation that meets both conditions. (E) First, find the midpoint of the line 2. 4. 3. ( )( ) 5. segment, which is where it intersects its perpendicular bisector. The midpoint’s xcoordinate is
4−3 1 = , and its y-coordinate is 2 2 The correct figure should show the graph of the equation y = x – 4. Choice (D) shows the graph of a line with slope 1 and y-intercept –4, which matches the features of this equation. No other answer choice provides a graph with both these features. 4. (D) Substitute (x+ 1) for x in the function: −2 + 5 3 = . Next, determine the slope of the 2 2 5 − (−2) 7 = = −1 . Since the line segment: −3 − 4 −7 f(x + 1) = [(x + 1) – 1]2 + 1 = x2 + 1 In the xy-plane, the equation of f(x + 1) is y = x2 + 1. To find the y-intercept of this equation’s graph, let x = 0, then solve for y: y = (0)2 + 1 = 1 5. (A) The graph of x = –...
View Full Document
- Spring '10