To determine. The correct graph for the function[g(x)=-frac{1}{2}f(x)+1 is B

To determine. The correct graph for the function[g(x)=-frac{1}{2}f(x)+1 is B

Question
Transformation properties
asked 2021-02-19
To determine.
The correct graph for the function\(\displaystyle{\left[{g{{\left({x}\right)}}}=-{\frac{{{1}}}{{{2}}}}{f{{\left({x}\right)}}}+{1}\right.}\) is B

Answers (1)

2021-02-20

Vertical translation:
For \(\displaystyle{b}\ {>}\ {0}\),
The graph of \(\displaystyle{y}={f{{\left({x}\right)}}}\ +\ {b}\) is the graph of \(\displaystyle{y}={f{{\left({x}\right)}}}\) {shifted up b units}
The graph of \(\displaystyle{y}={f{{\left({x}\right)}}}\ -\ {b}\) is the graph of \(\displaystyle{y}={f{{\left({x}\right)}}}\ +\ {b}\) {shifted down b units}
Reflection:
Across the x-axis:
The graph of \(\displaystyle{y}=\ -{f{{\left({x}\right)}}}\) is the reflection of the graph of \(\displaystyle{y}={f{{\left({x}\right)}}}\) across the x-axis
Across the y-axis:
The graph of \(\displaystyle{y}=\ {f{{\left(-{x}\right)}}}\) is the reflection of the graph of \(\displaystyle{y}={f{{\left({x}\right)}}}\) across the x-axis
Vertically Stretching or Shrinking:
The graph of \(\displaystyle{y}=\ {a}{f{{\left(-{x}\right)}}}\) can be obtained from the graph of \(\displaystyle{y}={f{{\left({x}\right)}}}\) by
Stretching verttically for \(\displaystyle{m}{i}{d}{a}{m}{i}{d}\ {>}\ {1}\) or shrinking vertically for \(\displaystyle{a}\ {>}\ {m}{i}{d}{a}{m}{i}{d}\ {>}\ {1}\).
For \(\displaystyle{a}{<}{0}\), the graph is also reflected across the x-axis
. By the properties of transformation, the graph of \(\displaystyle{g{{\left({x}\right)}}}=-{\frac{{{1}}}{{{2}}}}{f{{\left({x}\right)}}}+{1}\) is,
The transformation of \(\displaystyle{y}={f{{\left({x}\right)}}}\) and shri
vertically by a factor of \(\displaystyle{\frac{{{1}}}{{{2}}}}\)
Then the graph is reflection about the x-axis and shifted up one unit.
From the given graphs, the graphs B satisfies the all condition's stated above.
Thus, the correct graph for the function \(\displaystyle{g{{\left({x}\right)}}}=-{\frac{{{1}}}{{{2}}}}{f{{\left({x}\right)}}}+{1}\) is B

0

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