How does the graph of the transformed function x= − log3(5x −1)+2 compare to the graph of its parent function compare to the graph of its parent function f(x)= log3 x?
assuming base 3, we have
f(x) = logx
g(x) = -3log(5(x-1/5))+2
so need to transform f(x) as follows:
dilate horizontally by a factor of 1/5:
g(x) = log(5x)
shift right by 1/5:
g(x) = log(5(x - 1/5))
dilate vertically by a factor of 3:
g(x) = 3log(5(x - 1/5))
reflect across the x-axis:
g(x) = -3log(5(x - 1/5))
shift up 2:
g(x) = -3log(5(x - 1/5)) + 2
To compare the graph of the transformed function x = -log3(5x - 1) + 2 to the graph of its parent function f(x) = log3(x), we need to look at the effects of the transformations applied to the parent function. Let's break it down step by step:
1. Reflection about the x-axis:
The negative sign in front of the logarithm function (-log3) reflects the graph of the function over the x-axis. This means that any points on the parent function that are above the x-axis will be below the x-axis in the transformed function, and vice versa.
2. Horizontal compression:
The number 5 in the expression 5x inside the logarithm function is responsible for horizontal compression. It squeezes the graph horizontally, making it narrower. This means the transformed function will have a steeper slope than the parent function, as the compressed x-values will lead to a more rapid increase or decrease in the y-values (logarithms).
3. Horizontal translation:
The -1 in the expression 5x - 1 inside the logarithm function causes a horizontal shift to the left by 1 unit. This means all the x-values of the transformed function will be 1 unit to the left compared to the parent function.
4. Vertical translation:
The +2 outside the logarithm function causes a vertical shift upward by 2 units. This means all the y-values of the transformed function will be 2 units higher than the corresponding y-values of the parent function.
By combining these transformations, we can determine the relation between the two graphs. In summary, the transformed function x = -log3(5x - 1) + 2 will have the following characteristics in comparison to its parent function f(x) = log3(x):
- It will be a reflection of the parent function over the x-axis.
- The graph will be horizontally compressed, resulting in a steeper slope.
- It will be shifted 1 unit to the left compared to the parent function.
- The graph will be shifted 2 units higher vertically.
Keeping these transformations in mind, you can now visualize and compare the graphs of the transformed function and its parent function.
To understand how the graph of the transformed function x = -log3(5x - 1) + 2 compares to the graph of its parent function f(x) = log3 x, we need to analyze each transformation step by step.
1. Reflecting in the x-axis:
The negative sign in front of the logarithm function, x = -log3(5x - 1) + 2, reflects the graph in the x-axis. This means that the transformed function will be a mirror image of the parent function, f(x) = log3 x, with respect to the x-axis.
2. Horizontal compression/stretch:
The coefficient 5 in the argument of the logarithm, log3(5x - 1), determines the horizontal compression/stretch of the graph. Since the coefficient is greater than 1, the graph will be horizontally compressed. This means that the transformed function will be narrower than the parent function.
3. Horizontal translation:
The -1 within the argument of the logarithm, log3(5x - 1), represents a horizontal translation. This means that the graph of the transformed function will be shifted 1 unit to the right compared to the parent function.
4. Vertical translation:
The +2 outside the logarithm, x = -log3(5x - 1) + 2, represents a vertical translation. This means that the graph of the transformed function will be shifted 2 units upward compared to the parent function.
In summary, the graph of the transformed function x = -log3(5x - 1) + 2 will be a mirror image of the parent function f(x) = log3 x with respect to the x-axis. It will be horizontally compressed, shifted 1 unit to the right, and shifted 2 units upward.