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In y=(x−h)2+k, the graph of y=x2 shifts horizontally by h and vertically by k. The vertex becomes (h,k).
Remember A plus inside the bracket gives a shift to the left: (x + 3)² has vertex (−3, 0).
Worked example 1
01 / A point P=(−2,0) lies on y=f(x), whose domain is [−3,5] and range is [−2,7]. For g(x)=f(x)+(3), locate P’s image and give the new domain and range.
02 / Only the output changes: (u,w)→(u,w+3).
03 / The mapped point is (−2,3).
04 / Inputs remain [−3,5]; add 3 to both range endpoints.
Worked example 2 · Shift outputs vertically
A point P=(−1,1) lies on y=f(x), whose domain is [−3,5] and range is [−2,7]. For g(x)=f(x)+(−4), locate P’s image and give the new domain and range.
Only the output changes: (u,w)→(u,w−4). The mapped point is (−1,−3). Inputs remain [−3,5]; add −4 to both range endpoints.
Read the graphs
Recover transformation formulas from actual graphs
The blue graph is f(x)=x2. The orange graph is g, obtained by a horizontal shift followed by a positive vertical scale and vertical shift. Recover g in terms of f and give a corresponding explicit formula. Use the displayed scales and verify your rule algebraically.
The characteristic point ( vertex, endpoint, or inflection ) moves horizontally by 1 and vertically by 4. Compare an additional matched point: its change from the characteristic height is multiplied by 1. Check matching points using (u,w)→(u+1,1w+4); then substitute the stated f into g.
Answer: g(x)=f(x−1)+4=1(x−1)2+4
Transform unfamiliar graphs point by point
The graph is defined only on the shown closed polyline. Draw g(x)=2f(2(x−(−3)))+2. Label every transformed vertex and give its domain. Complete the written work or drawing first. The selected answer checks the result only; your reasoning or drawing needs teacher/self-review.
The point rule is (u,w)→(−3+2u,2w+2). Apply it to all four vertices: (−29,2), (−27,8), (−25,0), (−23,6). Join the corresponding consecutive vertices with straight segments; keep both endpoint dots closed. Written-work review:
Accurately read all original vertices from the SVG.
Correctly map all vertices and retain connectivity.
Draw closed endpoints and state the transformed domain.
This graph is given only for x≥0. Draw two separate full graphs: an even extension and an odd extension. State the output at x=−1 for each. Explain why the point at x=0 permits the odd extension. Complete the written work or drawing first. The selected answer checks the result only; your reasoning or drawing needs teacher/self-review.
For the even extension, map (u,w) to (−u,w). For the odd extension, map (u,w) to (−u,−w). At zero an odd function must satisfy f(0)=−f(0), hence f(0)=0, which the supplied graph does. Written-work review:
Two separately labeled constructions.
Correct y-axis reflection for even and origin reflection for odd.
Explain the necessary output zero at the origin.
Answer: Even output at −1: 3; odd output at −1: −3; f(0)=0 satisfies the odd condition.
Interpret transformations in real-world stories
The graph gives a walking group’s distance from its starting point ( meters ) versus time ( minutes ). Describe the original trip and calculate its outbound speed, then draw D(t)=(23)f(t) and G(t)=f(t−3). Give the new outbound speed for D and the new starting time for G. Complete the written work or drawing first. The selected answer checks the result only; your reasoning or drawing needs teacher/self-review.
Rising distance means outward travel, flat means stopped, falling means returning. Vertical scaling multiplies all distances and distance/time rates by (23) while preserving times. Replacing t with t−3 delays every event by 3 minutes; the pause duration remains 4 minutes. Written-work review:
Interpret all three stages.
Draw both transformed graphs with labeled times and distances.
Correct speed, delay, and unchanged pause duration.
Answer: Walk 22m in 5min at 522 meters per minute, pause 4min, return in 5min; D speed 533 meters per minute; G starts at 3min.
Textbook reference: Stewart, Redlin & Watson, Precalculus: Mathematics for Calculus, 7th ed. (Cengage, 2016). The explanations and practice here are original, organized by textbook section.