The figure on the right shows the graph of y = f(x) = -x.
Use the graph to sketch the graph of each of the following functions. Mark on each sketched graph the new position and coordinates of the point (2,-2).
(a) y = f(x + 4) + 3
(b) y = 2f(x) + 3

The answers to (a) and (b) are given as the two pioints (-2,1) and (2,-1) respectively.

I get stuck with the tropic of quadratic equation transformations, particularly about using a given graph to sketch a graph.

2. Originally Posted by Jon K

The figure on the right shows the graph of y = f(x) = -x.
Use the graph to sketch the graph of each of the following functions. Mark on each sketched graph the new position and coordinates of the point (2,-2).
(a) y = f(x + 4) + 3
(b) y = 2f(x) + 3

The answers to (a) and (b) are given as the two pioints (-2,1) and (2,-1) respectively.
Hello,

a) the use of the argument (x+4) means that you translate the graph of f by 4 units to the left. Adding 3 means that you translate the graph by 3 units up:

$(2, -2)\overbrace{~\longrightarrow~}^{\text{4 to left}}(2-4,-2) = (-2, -2) \overbrace{~\longrightarrow~}^{\text{3 up}}(-2, -2+3)=(-2, 1)$

b) the factor 2 doubles all values of f and adding 3 to these new values means that you translate the graph by 3 units upward:

$(2, -2)\overbrace{~\longrightarrow~}^{\text{double y-value}}(2,2 \cdot (-2)) = (2, -4) \overbrace{~\longrightarrow~}^{\text{3 up}}(-2, -4+3)=(-2, -1)$

3. Hello,

Thanks for your help. Really appreciated it. Your solution have helped me solve several additional questions in the assignment.

Could you please help me with this one about quadratic functions, too? I have spent quite some time on it but I can’t work out the two transformation rules. Many thanks.

(a) Use the given graph of the function y = (x + 2)² to sketch the graph of the new function y = (x – 3)² and mark on it the new position and coordinates of the point (-2, 0) after the transformation.

(b) The figure on the right shows the graph of the function y = x² + 1 and a point (1, 2) on it.
Sketch the new graph of y = (x – 2)² + 1 and mark on it the new position and coordinates of the point (1, 2) after the transformation.

The answers given are (3, 0) and (3, 2) respectively.

Thank again.
Jon K

4. Originally Posted by Jon K
...

(a) Use the given graph of the function y = (x + 2)² to sketch the graph of the new function y = (x – 3)² and mark on it the new position and coordinates of the point (-2, 0) after the transformation.

(b) The figure on the right shows the graph of the function y = x² + 1 and a point (1, 2) on it.
Sketch the new graph of y = (x – 2)² + 1 and mark on it the new position and coordinates of the point (1, 2) after the transformation.

The answers given are (3, 0) and (3, 2) respectively.
Hi,

to a) If the parabola y = x² is translated horizontally by h units to the right you'll get the equation
y = (x - h)²

Thus the equation y = (x+2)² indicates a translation by 2 units to the left.
An equation y = (x - 3)² indicates a translation by 3 units to the right. That means you have to translate the 2nd parabola by 5 units to the right:

$(-2, 0)\overbrace{~\longrightarrow~}^{\text{5 to right}}(-2+5, 0) = (3,0)$

to b) According the considerations at a) the replacement of x by (x-2) yields a translation 2 units to the right:

$(1, 2)\overbrace{~\longrightarrow~}^{\text{2 to right}}(1 + 2, 2) = (3, 2)$

You mentioned 2 transformation rules. I assume that you mean:

1. Horizontal translation by h units: $y = x^2~\longrightarrow~y = (x-h)^2$

2. Vertical translation by k units: $y=x^2~\longrightarrow~y=x^2+k$

You may collect both movements in one equation:

$y=(x-h)^2+k$

The parabola y = x² was translated by h units horizontally and by k units vertically.