Source: ACE Exam Paper 1 ADV
Tags: graph transformations, square root function, translations, horizontal shift, vertical shift, absolute value, modulus, quadratic inequality, graphical solutions
Graph transformations (shifts, stretches, reflections) come up both as standalone sketching questions and as tools for solving equations and inequalities. Knowing how y = f(x - h) + k moves the graph is fundamental. Absolute value graphs fold everything above the x-axis and create piecewise problems that often appear in "find m" or "find the number of solutions" questions.
Horizontal translation
y = f(x - h) shifts the graph h units to the right (if h > 0). The subtraction inside the function is the key indicator.
Vertical translation
y = f(x) + k shifts the graph k units up (if k > 0).
Absolute value function |f(x)|
Reflects any portion of f(x) that lies below the x-axis upward, so the output is always non-negative. Graphically, the negative parts "fold up."
These three functions illustrate horizontal and vertical translations of the basic square root curve.
y = √x
Domain: x ≥ 0
Starts at the origin (0, 0)
Standard square root shape: rises steeply at first, then flattens
y = √(x - 1)
The graph of y = √x shifted 1 unit to the right
Domain: x ≥ 1
Starts at (1, 0)
Same shape as y = √x, just repositioned
y = √x - 1
The graph of y = √x shifted 1 unit down
Domain: x ≥ 0
Starts at (0, -1)
Crosses the x-axis where √x = 1, i.e. at x = 1, so it passes through (1, 0)
All three pass through the point (1, 0) is not quite right: y = √x passes through (1, 1), y = √(x - 1) passes through (1, 0), and y = √x - 1 passes through (1, 0). So the two translated graphs share the point (1, 0), but the parent curve y = √x passes through (1, 1).
Consider y = x² and y = x² - 3x + 2 (which factors as (x - 1)(x - 2)).
The inequality x² > (x - 1)(x - 2) holds wherever the graph of y = x² is above y = x² - 3x + 2.
Set them equal: x² = x² - 3x + 2, giving 3x = 2, so x = 2/3.
For x < 2/3: test x = 0. LHS = 0, RHS = 2. So x² < (x - 1)(x - 2). The inequality does not hold.
For x > 2/3: test x = 1. LHS = 1, RHS = 0. So x² > (x - 1)(x - 2). The inequality holds.
Solution: x > 2/3.
The graph of y = |2x - 4|:
The expression inside is zero when 2x - 4 = 0, so x = 2
For x ≥ 2: y = 2x - 4 (a line with slope 2, y-intercept -4, but only the part from x = 2 onward)
For x < 2: y = -(2x - 4) = -2x + 4 (a line with slope -2)
The graph is a V-shape with vertex at (2, 0), opening upward
The left arm has slope -2 and the right arm has slope 2
For the equation |2x - 4| = mx + 1 to have exactly one solution, the line y = mx + 1 must touch the V-graph at exactly one point. The line passes through (0, 1) for all values of m.
There are three cases that give exactly one solution:
The line is tangent to one arm (parallel to one arm while intersecting the other at exactly one point). Since the arms have slopes 2 and -2, a line with slope 2 through (0, 1) hits the left arm once and runs parallel to the right arm (never touching it), giving one solution. Similarly for slope -2.
The line passes through the vertex (2, 0): 0 = 2m + 1, so m = -1/2. Check: y = -x/2 + 1 intersects the V at (2, 0). Since it is less steep than both arms, it touches at only one point on the way down.
The values of m for exactly one solution are m = 2, m = -2, and m = -1/2.
y = f(x - h): horizontal shift right by h
y = f(x) + k: vertical shift up by k
y = |f(x)|: reflect any negative portions of f(x) above the x-axis
|ax + b| = 0 when x = -b/a (the vertex of the V-graph)
⚠️ y = √(x - 1) is a horizontal shift (starts at x = 1). y = √x - 1 is a vertical shift (starts at y = -1). The brackets make all the difference.
⚠️ When sketching three related functions on the same axes, label each curve clearly and mark key points (start points, intercepts). Marks are often for showing these features.
⚠️ For |f(x)| = g(x), sketch both graphs and count intersections visually before doing algebra. The graph prevents you from missing solutions or inventing extras.
⚠️ "Exactly one solution" questions usually have multiple values of the parameter (here, m) that work. Identify all of them. Losing one means losing marks.
⚠️ When solving an inequality graphically, the intersection point is the boundary. Test a point on each side to confirm the direction of the inequality.
Q: How does y = √(x - 1) differ from y = √x?
A: It is the graph of y = √x shifted 1 unit to the right. The starting point moves from (0, 0) to (1, 0).
Q: Solve x² > (x - 1)(x - 2).
A: The graphs intersect at x = 2/3. Testing shows x² is above for x > 2/3.
Q: Sketch y = |2x - 4|. Where is the vertex?
A: V-shape with vertex at (2, 0). Left arm slope -2, right arm slope 2.
Q: For |2x - 4| = mx + 1 to have exactly one solution, what are the possible values of m?
A: m = 2, m = -2, or m = -1/2.
graph transformations, translations, horizontal shift, vertical shift, square root function, absolute value function, modulus function, V-graph, quadratic inequality, graphical solution, number of solutions, parameter problems, HSC maths advanced, Year 12 functions and graphs