Further Transformations of 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; multiplying by \(a\) stretches it. This Year 10A maths page applies them to parabolas, exponentials, circles and hyperbolas, and uses the transformed equations to find features, ranges 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 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.
These transformations are used in every senior mathematics course that works with graphs of functions.
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 \(y = 0\).
- Range of an exponential: \(b^{x} \gt 0\) for every \(x\), so \(y = b^{x} + c\) has range \(y \gt c\), and the reflection \(y = -b^{x} + c\) has range \(y \lt c\).
- 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) Write the equation of the new graph.
(ii) Find its \(x\)- and \(y\)-intercepts.
(iii) State the equation of its asymptote and its range.
(i) Reflecting in the \(x\)-axis changes the sign of every \(y\)-value: \(y = -3^{x}\). Then add \(9\) for \(9\) units up.
| \(y\) | \(=\) | \(\boldsymbol{-3^{x} + 9}\) |
(ii) \(y\)-intercept: put \(x = 0\). \(x\)-intercept: put \(y = 0\) and write \(9\) as a power of \(3\).
| \(x = 0:\quad y\) | \(=\) | \(-3^{0} + 9 = -1 + 9 = \mathbf{8}\) |
| \(y = 0:\quad 3^{x}\) | \(=\) | \(9 = 3^{2}\) |
| \(x\) | \(=\) | \(\mathbf{2}\) |
The intercepts are \((0,\ 8)\) and \((2,\ 0)\).
(iii) The asymptote of \(y = 3^{x}\) is \(y = 0\). The reflection keeps it at \(y = 0\), and the move up takes it to \(\mathbf{y = 9}\). Since \(3^{x} \gt 0\), \(-3^{x} \lt 0\), so \(y = -3^{x} + 9 \lt 9\). The range is \(\mathbf{y \lt 9}\).
The curve falls from left to right and gets closer and closer to \(y = 9\) on the left.
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.
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.