Fun House Mirror

Algebra Level 2

Given the graph y = ln x y = \ln x , which of these statements describes the transformations to get the graph of y = ln ( 4 x 2 + 4 x + 1 ) y = \ln (4x^2 + 4x + 1) for x > 1 2 x > - \frac{1}{2} ?

(1) \quad \text{(1)} Translate to the left by 1 and up by ln 4 \ln 4 , then scale vertically by 2.

(2) \quad \text{(2)} Translate to the left by 1 2 \frac{1}{2} and up by ln 2 \ln 2 , then scale vertically by 2.

(3) \quad \text{(3)} Translate to the left by 1 and up by ln 2 \ln 2 , then scale vertically by 2.

(4) \quad \text{(4)} Translate to the left by 1 2 \frac{1}{2} and up by ln 4 \ln 4 , then scale vertically by 2.

Statement number 1 Statement number 2 Statement number 3 Statement number 4

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2 solutions

Chung Kevin
Nov 24, 2014

Let us work through the options

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

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

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

The first transformation gives us y = ln ( x + 1 2 ) y = \ln ( x + \frac{1}{2} ) .
The second transformation gives us y = ln ( x + 1 2 ) + ln 2 y = \ln (x+ \frac{1}{2} ) + \ln 2 .
The third transformation gives us y = 2 [ ln ( x + 1 2 ) + ln 2 ] = ln ( 4 x 2 + 4 x + 1 ) 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 \ln 2 , scale the graph vertically by 2.

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

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

The first transformation gives us y = ln ( x + 1 2 ) y = \ln ( x + \frac{1}{2} ) .
The second transformation gives us y = ln ( x + 1 2 ) + ln 4 y = \ln (x+ \frac{1}{2} ) + \ln 4 .
The third transformation gives us y = 2 [ ln ( x + 1 2 ) + ln 4 ] = ln ( 16 x 2 + 16 x + 4 ) 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 1 2 \frac{1}{2} , translate the graph up by ln 2 \ln 2 , scale the graph vertically by 2.

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

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

Moderator note:

Thanks for the detailed explanation! This helps those who are not too familiar with graph transformations to understand what is going on.

Jakub Šafin
Nov 25, 2014

Take the point ( 1 , ln ( 1 ) ) = ( 1 , 0 ) (1,\ln(1))=(1,0) . If the first transformation is moving it to the left by 1, it wil be ( 0 , 0 ) (0,0) and the next two transformations have to keep it there, since the resulting y y for x = 1 x=1 is ln ( 1 ) = 0 \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 1/2 , then it's ( 1 / 2 , 0 ) (1/2,0) and the next two transformations have to make it ( 1 / 2 , ln ( 4 ) ) = ( 1 / 2 , 2 ln ( 2 ) ) (1/2,\ln(4))=(1/2,2\ln(2)) . Only one option accomplishes that, the other would make it ( 1 / 2 , 2 ln ( 4 ) ) (1/2,2\ln(4)) instead.

It helps to understand graphical transformations by analyzing them in a "pointwise" manner.

Chung Kevin - 6 years, 6 months ago

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