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To find the translation of the function [tex]\( f(x) = \frac{1}{x-2} - 1 \)[/tex] that results in the function [tex]\( h(x) = \frac{1}{x} \)[/tex], let's follow these steps:
1. Rewrite [tex]\( f(x) \)[/tex] in a more comparable form:
[tex]\[ f(x) = \frac{1}{x-2} - 1 \][/tex]
We can express the subtraction of 1 as a fraction over a common denominator:
[tex]\[ f(x) = \frac{1}{x-2} - \frac{x-2}{x-2} \][/tex]
Combining these terms gives:
[tex]\[ f(x) = \frac{1 - (x-2)}{x-2} = \frac{1 - x + 2}{x-2} = \frac{3 - x}{x-2} \][/tex]
2. Determine the shift needed to simplify the expression:
Looking at [tex]\( \frac{3 - x}{x-2} \)[/tex], we notice that we want an expression that matches [tex]\( h(x) = \frac{1}{x} \)[/tex]. This indicates a horizontal shift and possibly a modification within the fraction.
3. Apply the transformation to [tex]\( f(x) \)[/tex]:
To match [tex]\( h(x) = \frac{1}{x} \)[/tex], consider transforming [tex]\( x \)[/tex] such that the shift corrects the [tex]\( x-2 \)[/tex] in the denominator. Specifically, we add 2 to [tex]\( x \)[/tex]:
[tex]\[ \text{Let } x' = x + 2 \][/tex]
Substitute [tex]\( x' \)[/tex] back into the function [tex]\( f(x) \)[/tex]:
[tex]\[ f(x') = \frac{1}{x' - 2} - 1 = \frac{1}{(x + 2) - 2} - 1 = \frac{1}{x} - 1 \][/tex]
Now, to obtain [tex]\( h(x) \)[/tex], we need to adjust:
[tex]\[ f(x' - 2) + 1 = \frac{1}{x} - 1 + 1 = \frac{1}{x} \][/tex]
4. Verify the transformation results:
[tex]\[ f(x+2) + 1 = \frac{1}{(x+2)-2} - 1 + 1 = \frac{1}{x} \][/tex]
Thus, we have:
[tex]\[ \left( \frac{1}{x-2} - 1 \right) \text{ shifted horizontally by 2 units and added by 1 results in } \frac{1}{x} \][/tex]
Therefore, to transform [tex]\( f(x) = \frac{1}{x-2} - 1 \)[/tex] into [tex]\( h(x) = \frac{1}{x} \)[/tex], you horizontally shift [tex]\( f(x) \)[/tex] by 2 units to get rid of the [tex]\(-2\)[/tex] in the denominator and then add 1 vertically to cancel out the [tex]\(-1\)[/tex] term. This gives:
[tex]\( f(x+2) + 1 = \frac{1}{x} \)[/tex]
This completes the translation from [tex]\( f(x) \)[/tex] to [tex]\( h(x) \)[/tex].
1. Rewrite [tex]\( f(x) \)[/tex] in a more comparable form:
[tex]\[ f(x) = \frac{1}{x-2} - 1 \][/tex]
We can express the subtraction of 1 as a fraction over a common denominator:
[tex]\[ f(x) = \frac{1}{x-2} - \frac{x-2}{x-2} \][/tex]
Combining these terms gives:
[tex]\[ f(x) = \frac{1 - (x-2)}{x-2} = \frac{1 - x + 2}{x-2} = \frac{3 - x}{x-2} \][/tex]
2. Determine the shift needed to simplify the expression:
Looking at [tex]\( \frac{3 - x}{x-2} \)[/tex], we notice that we want an expression that matches [tex]\( h(x) = \frac{1}{x} \)[/tex]. This indicates a horizontal shift and possibly a modification within the fraction.
3. Apply the transformation to [tex]\( f(x) \)[/tex]:
To match [tex]\( h(x) = \frac{1}{x} \)[/tex], consider transforming [tex]\( x \)[/tex] such that the shift corrects the [tex]\( x-2 \)[/tex] in the denominator. Specifically, we add 2 to [tex]\( x \)[/tex]:
[tex]\[ \text{Let } x' = x + 2 \][/tex]
Substitute [tex]\( x' \)[/tex] back into the function [tex]\( f(x) \)[/tex]:
[tex]\[ f(x') = \frac{1}{x' - 2} - 1 = \frac{1}{(x + 2) - 2} - 1 = \frac{1}{x} - 1 \][/tex]
Now, to obtain [tex]\( h(x) \)[/tex], we need to adjust:
[tex]\[ f(x' - 2) + 1 = \frac{1}{x} - 1 + 1 = \frac{1}{x} \][/tex]
4. Verify the transformation results:
[tex]\[ f(x+2) + 1 = \frac{1}{(x+2)-2} - 1 + 1 = \frac{1}{x} \][/tex]
Thus, we have:
[tex]\[ \left( \frac{1}{x-2} - 1 \right) \text{ shifted horizontally by 2 units and added by 1 results in } \frac{1}{x} \][/tex]
Therefore, to transform [tex]\( f(x) = \frac{1}{x-2} - 1 \)[/tex] into [tex]\( h(x) = \frac{1}{x} \)[/tex], you horizontally shift [tex]\( f(x) \)[/tex] by 2 units to get rid of the [tex]\(-2\)[/tex] in the denominator and then add 1 vertically to cancel out the [tex]\(-1\)[/tex] term. This gives:
[tex]\( f(x+2) + 1 = \frac{1}{x} \)[/tex]
This completes the translation from [tex]\( f(x) \)[/tex] to [tex]\( h(x) \)[/tex].
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