Fun House Mirror

Let us work through the options

1) Translate the graph to the left by 1, translate the graph up by $\ln 4$ , scale the graph vertically by 2.

The first transformation gives us $y = \ln (x+1)$ .
The second transformation gives us $y = \ln (x+1) + \ln 4$ .
The third transformation gives us $y = 2 [ \ln (x+1) + \ln 4] = \ln 16 ( x^2 + 2x + 1 )$ .

2) Translate the graph to the left by $\frac{1}{2}$ , translate the graph up by $\ln 2$ , scale the graph vertically by 2.

The first transformation gives us $y = \ln ( x + \frac{1}{2} )$ .
The second transformation gives us $y = \ln (x+ \frac{1}{2} ) + \ln 2$ .
The third transformation gives us $y = 2 [ \ln (x+\frac{1}{2}) + \ln 2 ] = \ln (4 x^2 + 4x + 1 )$ .

3) Translate the graph to the left by 1, translate the graph up by $\ln 2$ , scale the graph vertically by 2.

The first transformation gives us $y = \ln ( x + 1 )$ .
The second transformation gives us $y = \ln (x + 1) + \ln 2$ .
The third transformation gives us $y = 2 [ \ln (x+1) + \ln 2 ] = \ln ( 4x^2 + 8x + 4 )$ .

4) Translate the graph to the left by $\frac{1}{2}$ , translate the graph up by $\ln 4$ , scale the graph vertically by 2.

The first transformation gives us $y = \ln ( x + \frac{1}{2} )$ .
The second transformation gives us $y = \ln (x+ \frac{1}{2} ) + \ln 4$ .
The third transformation gives us $y = 2 [ \ln (x+\frac{1}{2}) + \ln 4 ] = \ln (16 x^2 + 16x + 4 )$ .

Hence, the answer is 2) Translate the graph to the left by $\frac{1}{2}$ , translate the graph up by $\ln 2$ , scale the graph vertically by 2.

Note: For the full graph of $y = \ln ( 4x^2 + 4x + 1 )$ , we will also need to reflect it with respect to the line $x = - \frac{ 1}{2}$ . The reason for this is that

$y = \ln (4x^2 + 4x + 1) = 2 [ \ln 2 + \ln | x + \frac{1}{2}| ].$

Take the point $(1,\ln(1))=(1,0)$ . If the first transformation is moving it to the left by 1, it wil be $(0,0)$ and the next two transformations have to keep it there, since the resulting $y$ for $x=1$ is $\ln(1)=0$ . That can't work with either of the two options we have now.

If it's moving to the left just by $1/2$ , then it's $(1/2,0)$ and the next two transformations have to make it $(1/2,\ln(4))=(1/2,2\ln(2))$ . Only one option accomplishes that, the other would make it $(1/2,2\ln(4))$ instead.