AI HL · 12.5
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Translations of graphs
12.5.5
The dashed graph is y = f(x). The orange graph is y = p f(qx + c) + b. Drag the first orange point to translate the graph, or choose a kind of change and move the sliders. Say what happens to the labelled points before you read the answer.
Move a slider to change the graph.
The graph of \(y=f(x)\) is a downward parabola with its maximum at \((2, 5)\) and y-intercept 1, as shown. Sketch \(y=f(x)-3\) and give its maximum point and y-intercept.
Have a go first. Then open the working one line at a time.
\(y=f(x)-3\)
3 is taken from every output. The graph moves down 3. Nothing happens to x.
\((2, 5)\to(2, 2)\) \((0, 1)\to(0, -2)\)
Keep each x-value. Take 3 from each y-value.
Maximum (2, 2), y-intercept −2, same shape.
The dashed graph is the original moved down 3.
The graph of \(y=f(x)\) is made of straight lines through A(−3, −1), B(0, 3), C(3, −2) and D(5, 1). Sketch \(y=f(x+2)\). Give the image of C.
Have a go first. Then open the working one line at a time.
\(y=f(x+2)\)
The change is with x, so it acts on the x-values, and it works backwards: +2 moves the graph 2 units left.
Take 2 from each x-value: A'(−5, −1), B'(−2, 3), C'(1, −2), D'(3, 1)
Keep every y-value.
C' is (1, −2).
The dashed graph is the original moved 2 left.
The graph of \(y=x^2-5x+2\) is translated 3 units to the left. Write the new equation in the form \(y=ax^2+bx+c\).
Have a go first. Then open the working one line at a time.
\(y=f(x+3)=(x+3)^2-5(x+3)+2\)
Replace every x by (x + 3).
\(y=x^2+6x+9-5x-15+2\)
Expand (x + 3)² and −5(x + 3) separately.
\(y=x^{2} + x - 4\)
Collect like terms.
The graph of \(y=2x^3\) is translated 4 units to the right and 5 units down. Write down the new equation, and find y when x = 5.
Have a go first. Then open the working one line at a time.
\(y=2(x-4)^3\)
Right by 4 means replace x by x − 4.
\(y=2(x-4)^3-5\)
Down 5 means take 5 from the output.
\(y=2(5-4)^3-5=-3\)
Put in x = 5.