Further Transformations
Further transformations apply the same moves, translations and reflections, to every type of graph: parabolas, exponentials, hyperbolas, circles and cubics. Replacing \(x\) with \(x-h\) shifts a graph \(h\) units right, and adding \(k\) shifts it up. In NSW Year 10 Advanced maths you write new equations and find new asymptotes, centres and turning points (NESA outcome MA5‑NLI‑P‑01).
You will translate \(y=2^x\), \(y=x^2\), circles and hyperbolas, write the new equations, and state asymptotes, vertices and centres. You will find the equation of a hyperbola or circle from its graph, sketch \(y=-(x+1)^3-1\) and \(y=-3^x+9\), and find where lines meet parabolas, hyperbolas and circles. Harder questions translate a hyperbola so that it passes through the origin. It is part of Year 10 Advanced Non Linear Relationships Functions and Graphs.
Theory
The same rules transform every kind of graph. Replacing \(x\) by \(x - h\) moves a graph \(h\) units right; adding \(k\) moves it \(k\) units up; \(y \to -y\) and \(x \to -x\) reflect it in the axes. This Year 10 Advanced maths page applies them to parabolas, exponentials, circles, hyperbolas and cubics, and uses the transformed equations to find features and intersections.
A translation slides a graph without turning or resizing it. A reflection flips it over an axis. Both keep the shape; only the position (and, for a reflection, the orientation) changes.
Each family has a key feature that tracks the move: the vertex of a parabola, the centre of a circle, the point of inflection of a cubic, the horizontal asymptote of an exponential, and both asymptotes of a hyperbola.
Recognise the family from the equation: \(x^2\) only (no \(y^2\)) is a parabola; \(x^2\) and \(y^2\) with equal coefficients is a circle; \(x\) in an index is an exponential; \(x\) in a denominator is a hyperbola; a cube is a cubic.
It comes up if you are heading for Advanced or Extension maths in Years 11 and 12.
The rules, written for any graph \(y = f(x)\):
| Equation | Transformation | Point \((x,\ y)\) goes to |
|---|---|---|
| \(y = f(x - h)\) | \(h\) units right | \((x + h,\ y)\) |
| \(y = f(x) + k\) | \(k\) units up | \((x,\ y + k)\) |
| \(y = -f(x)\) | reflection in the \(x\)-axis | \((x,\ -y)\) |
| \(y = f(-x)\) | reflection in the \(y\)-axis | \((-x,\ y)\) |
The translated forms of each family (\(h\) right, \(k\) up):
How to transform a graph's equation
- Horizontal move: replace every \(x\) by \((x - h)\) for \(h\) right, or \((x + h)\) for \(h\) left.
- Vertical move: add \(k\) to the whole right-hand side (or replace \(y\) by \(y - k\) in a circle).
- Reflection: \(y \to -y\) for the \(x\)-axis, \(x \to -x\) for the \(y\)-axis.
- Track the key feature to check: it must move the same way.
Problem types
- Sketching: identify the family, draw the moved key feature (dashed asymptotes, centre dot, vertex), then intercepts from \(x = 0\) and \(y = 0\).
- Which equation fits? Eliminate options using the asymptotes, vertex or centre, then the direction (rising or falling, concave up or down).
- From a graph to an equation: read the asymptotes or centre to get \(h\) and \(k\); substitute a point for the last constant (Example 3).
- Sequences of moves: apply them in order. Moving \(y = \dfrac{5}{x + 1} - 2\) a further \(2\) right gives \(y = \dfrac{5}{(x - 2) + 1} - 2 = \dfrac{5}{x - 1} - 2\).
- Intersections: substitute the line into the curve. For a circle, it often helps to notice \(y + k\) or \(x - h\) simplifies (Example 4). For a hyperbola, multiply by the denominator. For a parabola, collect terms into a quadratic.
- Conditions with letters: if a translated graph must pass through a point, or its centre must lie on a line, substitute to get equations in \(h\) and \(k\) and solve them.
(i) \(y = x^2\)
(ii) \(y = 2^x\)
(iii) \(x^2 + y^2 = 4\)
(iv) \(y = \dfrac{1}{x}\).
\(4\) left: replace every \(x\) by \((x + 4)\). \(3\) up: replace \(y\) by \((y - 3)\), which for \(y = \ldots\) is the same as adding \(3\) to the right-hand side.
| Graph | New equation | Key feature moves to |
|---|---|---|
| (i) parabola | \(y = (x + 4)^2 + 3\) | vertex \((-4,\ 3)\) |
| (ii) exponential | \(y = 2^{x + 4} + 3\) | asymptote \(y = 3\) |
| (iii) circle | \((x + 4)^2 + (y - 3)^2 = 4\) | centre \((-4,\ 3)\), radius \(2\) |
| (iv) hyperbola | \(y = \dfrac{1}{x + 4} + 3\) | asymptotes \(x = -4\), \(y = 3\) |
Every point \((x,\ y)\) moves to \((x - 4,\ y + 3)\). Shape and size do not change: the circle keeps radius \(2\).
(i) in the \(x\)-axis
(ii) in the \(y\)-axis. Find each new equation and point of inflection.
(i) In the \(x\)-axis, every \(y\)-value changes sign: replace \(y\) by \(-y\).
| \(-y\) | \(=\) | \((x - 1)^3 + 2\) |
| \(y\) | \(=\) | \(\boldsymbol{-(x - 1)^3 - 2}\) |
Point of inflection \((1,\ 2) \to \mathbf{(1,\ -2)}\).
(ii) In the \(y\)-axis, every \(x\)-value changes sign: replace \(x\) by \(-x\).
| \(y\) | \(=\) | \((-x - 1)^3 + 2\) |
| \(\) | \(=\) | \(\left[-(x + 1)\right]^3 + 2\) |
| \(\) | \(=\) | \(\boldsymbol{-(x + 1)^3 + 2}\) |
Point of inflection \((1,\ 2) \to \mathbf{(-1,\ 2)}\). Since \((-a)^3 = -a^3\), the minus sign can come out of the cube.
Moving the asymptotes from \(x = 0,\ y = 0\) to \(x = 4,\ y = -1\) is a translation \(4\) right and \(1\) down: \(y = \dfrac{k}{x - 4} - 1\). Substitute \((0,\ 0)\):
| \(0\) | \(=\) | \(\dfrac{k}{-4} - 1\) |
| \(\dfrac{k}{-4}\) | \(=\) | \(1\) |
| \(k\) | \(=\) | \(-4\) |
So \(y = \boldsymbol{-\dfrac{4}{x - 4} - 1}\). Check \((2,\ 1)\):
| \(-\dfrac{4}{2 - 4} - 1\) | \(=\) | \(2 - 1 = 1\ \checkmark\) |
New equation: \((x - 2)^2 + (y + 1)^2 = 10\). Substitute \(y = x - 1\), so \(y + 1 = x\):
| \((x - 2)^2 + x^2\) | \(=\) | \(10\) |
| \(2x^2 - 4x + 4\) | \(=\) | \(10\) |
| \(x^2 - 2x - 3\) | \(=\) | \(0\) |
| \((x - 3)(x + 1)\) | \(=\) | \(0\) |
\(x = 3\) gives \(y = 2\); \(x = -1\) gives \(y = -2\). The line meets the circle at \(\mathbf{(3,\ 2)}\) and \(\mathbf{(-1,\ -2)}\).
Common pitfalls
Frequently asked questions
How do you translate a graph left or right?
Replace every x in the equation with (x - h) to move it h units right, or (x + h) to move it h units left. The change happens inside, to x, and works the opposite way to the sign.
How do you translate a graph up or down?
Add k to the whole right-hand side of y = f(x) to move it k units up, or subtract to move it down. For a circle, replace y with (y - k) instead.
How do you reflect a graph in the x-axis or y-axis?
To reflect in the x-axis, change the sign of y, giving y = -f(x). To reflect in the y-axis, change the sign of x, giving y = f(-x).
How do you tell which type of graph an equation makes?
x squared without y squared is a parabola; x squared and y squared with equal coefficients is a circle; x in an index is an exponential; x in a denominator is a hyperbola; x cubed is a cubic.
What happens to the asymptotes when a hyperbola is translated?
They move with it. Translating y = k over x by h right and c up moves the asymptotes from x = 0 and y = 0 to x = h and y = c.