The graph of y = log(-x) was transformed to y = -3log(2x + 4) – 1. Where did the point (10, 1) move to?
The Correct Answer and Explanation is:
The given function undergoes several transformations. Let’s break them down step-by-step to see how the point (10, 1) moves.
Original Function:
The original graph is given by y=log(x)y = \log(x)y=log(x), which is a logarithmic function with a vertical asymptote at x=0x = 0x=0 and passes through the point (1, 0).
Transformed Function:
The transformed function is y=−3log(2x+4)−1y = -3 \log(2x + 4) – 1y=−3log(2x+4)−1. To analyze the transformation, we must consider each operation one by one:
- Horizontal Shift:
y=log(2x+4)y = \log(2x + 4)y=log(2x+4) is the result of shifting y=log(x)y = \log(x)y=log(x) horizontally. To find the horizontal shift, set 2x+4=02x + 4 = 02x+4=0. Solving for xxx, we get: 2x=−4⇒x=−2.2x = -4 \quad \Rightarrow \quad x = -2.2x=−4⇒x=−2. This means the graph of y=log(2x+4)y = \log(2x + 4)y=log(2x+4) is shifted 2 units to the left. - Vertical Stretch and Reflection:
The term −3-3−3 outside the logarithmic function causes a vertical stretch by a factor of 3 and a reflection across the x-axis. So, all y-values are multiplied by -3. - Vertical Shift:
The term −1-1−1 at the end of the function shifts the entire graph 1 unit downward.
Transformation of the Point (10, 1):
To see where the point (10, 1) moves, substitute x=10x = 10x=10 into the transformed function:y=−3log(2(10)+4)−1=−3log(20+4)−1=−3log(24)−1.y = -3 \log(2(10) + 4) – 1 = -3 \log(20 + 4) – 1 = -3 \log(24) – 1.y=−3log(2(10)+4)−1=−3log(20+4)−1=−3log(24)−1.
Now, calculate log(24)\log(24)log(24). Using the approximation log(24)≈1.3802\log(24) \approx 1.3802log(24)≈1.3802:y=−3(1.3802)−1=−4.1406−1=−5.1406.y = -3(1.3802) – 1 = -4.1406 – 1 = -5.1406.y=−3(1.3802)−1=−4.1406−1=−5.1406.
Thus, the point (10, 1) moves to approximately (10, -5.14).
Conclusion:
After the transformations, the point (10,1)(10, 1)(10,1) moves to (10,−5.14)(10, -5.14)(10,−5.14).
