Translations Of Functions
Understand translations of functions for Queensland Year 11 Mathematical Methods (QCAA). A translation slides a graph across the plane without changing its shape or size, shifting it left, right, up or down.
You will learn to write the image rule for a horizontal and vertical shift, seeing the effect of the parameters h and k, map key points, and read off the new turning point, asymptote or intercept.
Every question with a fully worked solution.
- Translations Of Functions - Video - Translating Graphs of Functions Watch
Theory
In Year 11 Mathematical Methods (QCAA), a translation slides a graph without changing its shape. The image of \(y=f(x)\) after moving \(h\) units right and \(k\) units up is \(y=f(x-h)+k\). This page shows how to write the image rule, apply it to key points, and read off the new turning point, asymptote or intercept.
A translation is a transformation that shifts every point of a graph the same distance in the same direction; the curve keeps its size and shape. A vertical translation of \(k\) units gives \(y=f(x)+k\) (up when \(k\gt 0\), down when \(k\lt 0\)).
A horizontal translation of \(h\) units gives \(y=f(x-h)\). Watch the sign: replacing \(x\) with \(x-h\) moves the graph \(h\) units to the right, so a shift to the left uses \(x-(-c)=x+c\). Combining both, the image of \(y=f(x)\) is \(y=f(x-h)+k\).
Image of \(y=f(x)\) under a translation of \(h\) right and \(k\) up:
Effect on a point:
How to translate a graph \(h\) right and \(k\) up
- Read off \(h\) and \(k\): right/left gives \(h\) (right positive), up/down gives \(k\) (up positive).
- Write the rule: replace \(x\) with \(x-h\) and add \(k\) to get \(y=f(x-h)+k\).
- Move the key points: send each feature \((x,\,y)\) to \((x+h,\,y+k)\) — turning point, endpoint, asymptote or intercept.
Rule — add \(k=3\) to the function:
| \(y\) | \(=\) | \(f(x)+k\) |
| \(=\) | \(x^2+3\) |
Point — map \((x,\,y)\to(x,\,y+3)\):
| \((1,\,1)\) | \(\mapsto\) | \((1,\ 1+3)\) |
| \(=\) | \((1,\,4)\) |
Image rule \(y=x^2+3\); the point \((1,\,1)\) maps to \((1,\,4)\).
Rule — replace \(x\) with \(x-h\), where \(h=2\):
| \(y\) | \(=\) | \(f(x-h)\) |
| \(=\) | \((x-2)^2\) |
Right means a positive \(h\), so we subtract \(2\) inside the bracket.
Vertex — the turning point \((0,\,0)\) maps to \((0+2,\ 0)\):
| \((0,\,0)\) | \(\mapsto\) | \((2,\,0)\) |
Image rule \(y=(x-2)^2\); the vertex moves to \((2,\,0)\).
Read the shifts:
| \(h\) | \(=\) | \(1\) |
| \(k\) | \(=\) | \(2\) |
Rule — use \(y=f(x-h)+k\):
| \(y\) | \(=\) | \(f(x-1)+2\) |
| \(=\) | \(\sqrt{x-1}+2\) |
Endpoint — \((0,\,0)\) maps to \((0+1,\ 0+2)\):
| \((0,\,0)\) | \(\mapsto\) | \((1,\,2)\) |
Image rule \(y=\sqrt{x-1}+2\); the endpoint is \((1,\,2)\).
Match to \(y=f(x-h)+k\):
| \(x-h\) | \(=\) | \(x+3\) |
| \(h\) | \(=\) | \(-3\) |
| \(k\) | \(=\) | \(-4\) |
Since \(h=-3\) the shift is \(3\) units left; since \(k=-4\) it is \(4\) units down.
Turning point — \((0,\,0)\) maps to \((0-3,\ 0-4)\):
| \((0,\,0)\) | \(\mapsto\) | \((-3,\,-4)\) |
Translation \(3\) left and \(4\) down; turning point \((-3,\,-4)\).
Common pitfalls
Frequently asked questions
What does y = f(x - h) + k mean?
It is the image of \(y=f(x)\) after a translation of \(h\) units right and \(k\) units up. Every point \((x,\,y)\) moves to \((x+h,\,y+k)\).
Why does f(x - 2) shift the graph right, not left?
To get the same output the input must be \(2\) larger, so each point is reached \(2\) units further along the \(x\)-axis — the graph slides \(2\) units to the right.
How does a translation change the turning point of a parabola?
The turning point moves by the same translation: for \(y=(x-h)^2+k\) the vertex of \(y=x^2\) at \((0,\,0)\) moves to \((h,\,k)\).
Does a translation change the shape of the graph?
No. A translation only slides the graph; its size, orientation and width are unchanged, so \(y=x^2\) and \(y=(x-h)^2+k\) are congruent parabolas.
How do I translate the reciprocal graph y = 1/x?
Use \(y=\dfrac{1}{x-h}+k\). The vertical asymptote moves to \(x=h\) and the horizontal asymptote moves to \(y=k\).