Graph Transformations

An addition outside the function changes the y-coordinate.

1. The graph moves up 2 units.
2. Keep x=-1 and add 2 to the y-coordinate, giving (-1,6).

A subtraction outside the function changes the y-coordinate.

1. The graph moves down 3 units.
2. The corresponding point is (2,2-3)=(2,-1).

A subtraction inside the function moves the graph in the positive horizontal direction.

1. The graph moves right 4 units.
2. The corresponding point is (-3+4,1)=(1,1).
3. Track the point through the transformation.

An addition inside the function moves the graph left.

1. The graph moves left 2 units.
2. The corresponding point is (4-2,-1)=(2,-1).
3. Track the point through the transformation.

The negative sign outside the function changes the sign of the output.

1. Reflection in the x-axis leaves the x-coordinate unchanged.
2. Change the sign of the y-coordinate to get (2,-2).
3. Track the point through the transformation.

A negative input reflects the graph in the y-axis.

1. Reflection in the y-axis changes the sign of the x-coordinate.
2. The corresponding point is (3,1).
3. Track the point through the transformation.

Compare the x- and y-coordinates of matching corners.

1. Every orange corner is 3 units above the matching blue corner.
2. An upward translation is written g(x)=f(x)+3.

Track one matching corner and check the others move in the same way.

1. Each corner of the orange graph is 2 units to the left of the matching blue corner, at the same height.
2. Replacing x by x+2 moves a graph 2 units to the left, so g(x)=f(x+2).
3. f(x-2) would move it 2 units to the right: the sign inside the bracket works the opposite way to what you might expect.

Compare the signs of the y-coordinates at the same value of x.

1. The x-coordinates are unchanged and every y-coordinate has its sign reversed.
2. The formula is y=-f(x).

Check whether the x-coordinates or the y-coordinates change sign.

1. Each orange corner is the mirror image of a blue corner in the y-axis: the x-coordinate changes sign and the y-coordinate stays the same.
2. A reflection in the y-axis is y=f(-x).
3. y=-f(x) would reflect in the x-axis instead, turning the graph upside down.

Deal with the change inside the bracket and the change outside it separately.

1. f(x+4) moves the graph 4 units to the left: (3,-2) moves to (-1,-2).
2. Adding 1 outside the function moves the graph 1 unit up: (-1,-2) moves to (-1,-1).
3. The turning point is (-1,-1).
4. Track the point through the transformations.

The horizontal translation changes each root by the same amount.

1. The graph moves right 3 units.
2. Add 3 to each root: -4+3=-1 and 2+3=5.
3. Each root shifts three units to the right.

4. The new roots are -1 and 5.

Find the turning point of y=(x+1)^2+2, then move it 3 units right and 5 units down.

1. y=(x+1)^2+2 has its turning point where the bracket is 0: at (-1,2).
2. Move it 3 right and 5 down: (-1+3,\ 2-5)=(2,-3).
3. The marked turning points locate the original and transformed curves.

4. The new graph is the same shape with turning point (2,-3), so its equation is y=(x-2)^2-3.
5. Check with function notation: replace x by x-3 and subtract 5: (x-3+1)^2+2-5=(x-2)^2-3.

A reflection in the x-axis changes the sign of every y-value.

1. Reflecting in the x-axis changes the sign of every y-value, so the new graph is y=-(x^2-2).
2. Multiply out the bracket: y=-x^2+2.
3. The marked turning points locate the original and transformed curves.

4. Check: the lowest point (0,-2) becomes the highest point (0,2).

A point on the x-axis has y-coordinate 0.

1. y=-f(x) is y=f(x) reflected in the x-axis: every y-value changes sign and every x-value stays the same.
2. Where f(x)=0, -f(x)=-0=0 too, so the points on the x-axis do not move.
3. Points on the x-axis remain on the x-axis under this reflection.

4. So the graph crosses the x-axis at the same points. Changing the sign of the crossings would be y=f(-x), a reflection in the y-axis.

Reflect in the y-axis, then apply the translation upwards.

1. The reflection in the y-axis maps (-3,1) to (3,1).
2. Adding 3 to the output maps this to (3,4).
3. Track the point through the transformations.

Complete the square to find the turning point of C, then move it.

1. Complete the square: x^2-6x+5=(x-3)^2-9+5=(x-3)^2-4, so C has turning point (3,-4).
2. The vector \begin{pmatrix}-2\\5\end{pmatrix} moves every point 2 left and 5 up: (3,-4) moves to (1,1).
3. The marked turning points locate the original and transformed curves.

4. S is the same shape with turning point (1,1), so S is y=(x-1)^2+1.
5. Multiply out: (x-1)^2+1=x^2-2x+1+1, so S is y=x^2-2x+2.

Apply the horizontal shift and the output reflection separately.

1. The input change moves the graph left 1 unit, giving (-3,-4).
2. The minimum moves one unit left; its height stays the same.

3. Negating the output reflects this in the x-axis, giving (-3,4).
4. Reflection keeps the x-coordinate and changes the sign of the y-coordinate.

5. This turning point is now a maximum.

A change inside the bracket moves the graph left or right; a change outside moves it up or down.

1. \cos(x+60)^\circ is \cos x^\circ with x replaced by x+60, which moves the graph 60 to the left: (180,-1) moves to (120,-1).
2. Subtracting 2 outside moves the graph 2 down: (120,-1) moves to (120,-3).
3. Check: at x=120, \cos(120+60)^\circ-2=\cos180^\circ-2=-3, the lowest value of the new graph.

A reflection in the y-axis is y=f(-x): replace every x with -x.

Part (a)

1. Reflecting in the y-axis gives y=f(-x), so replace every x with -x.
2. (-x)^2-4(-x)+1=x^2+4x+1, so D is y=x^2+4x+1.
3. The two curves are mirror images across the y-axis.

Part (b)

1. Complete the square: x^2+4x+1=(x+2)^2-4+1=(x+2)^2-3.
2. The turning point of D is (-2,-3).
3. Check: C is (x-2)^2-3, with turning point (2,-3); reflecting in the y-axis changes the sign of the x-coordinate.
4. The turning point keeps its y-coordinate, and its x-coordinate changes sign.