Answer :
To analyze how the features of the graphs of the functions [tex]\( g(x) = -f(x) + 10 \)[/tex] and [tex]\( h(x) = -f(x-4) + 1 \)[/tex] differ from the original function [tex]\( f(x) \)[/tex], we need to consider the effects of the transformations applied to [tex]\( f(x) \)[/tex]. Here's a step-by-step breakdown:
1. Range:
- [tex]\( g(x) = -f(x) + 10 \)[/tex]:
- The "-f(x)" part reflects the graph of [tex]\( f(x) \)[/tex] over the x-axis, effectively flipping the range. Therefore, if [tex]\( f(x) \)[/tex] has values from [tex]\( a \)[/tex] to [tex]\( b \)[/tex], after reflection, those values become [tex]\( -a \)[/tex] to [tex]\( -b \)[/tex].
- The "+10" part shifts the entire graph up by 10 units. This means each value in the range will increase by 10 units. Therefore, the new range will be from [tex]\( -a + 10 \)[/tex] to [tex]\( -b + 10 \)[/tex].
- [tex]\( h(x) = -f(x-4) + 1 \)[/tex]:
- The "-f(x-4)" part first shifts the graph of [tex]\( f(x) \)[/tex] four units to the right and then reflects it over the x-axis, thus flipping the range.
- The "+1" part then shifts the graph up by 1 unit, resulting in an adjusted range from [tex]\( -a + 1 \)[/tex] to [tex]\( -b + 1 \)[/tex].
Thus, the range of both [tex]\( g(x) \)[/tex] and [tex]\( h(x) \)[/tex] will differ from [tex]\( f(x) \)[/tex].
2. Domain:
- Neither vertical reflections nor shifts (horizontal or vertical) affect the domain of the function. The domain of [tex]\( g(x) \)[/tex] and [tex]\( h(x) \)[/tex] remains the same as [tex]\( f(x) \)[/tex].
3. y-intercept:
- For [tex]\( g(x) = -f(x) + 10 \)[/tex]:
- The y-intercept of [tex]\( f(x) \)[/tex] is affected by both the reflection and the vertical shift. If [tex]\( f(0) = c \)[/tex], then the y-intercept of [tex]\( g(x) \)[/tex] will be [tex]\( -c + 10 \)[/tex].
- For [tex]\( h(x) = -f(x-4) + 1 \)[/tex]:
- The y-intercept of [tex]\( h(x) \)[/tex] is found by evaluating [tex]\( h(0) \)[/tex], which is [tex]\( -f(-4) + 1 \)[/tex]. This shifts the y-intercept to the point [tex]\( -f(-4) + 1 \)[/tex].
Therefore, the y-intercept for both [tex]\( g(x) \)[/tex] and [tex]\( h(x) \)[/tex] is different from that of [tex]\( f(x) \)[/tex].
4. End behavior:
- [tex]\( g(x) = -f(x) + 10 \)[/tex]:
- The end behavior of [tex]\( f(x) \)[/tex] is first reflected about the x-axis and then shifted up by 10 units.
- [tex]\( h(x) = -f(x-4) + 1 \)[/tex]:
- The end behavior of [tex]\( f(x) \)[/tex] is first shifted right by 4 units, reflected, and then shifted up by 1 unit.
The end behavior of both [tex]\( g(x) \)[/tex] and [tex]\( h(x) \)[/tex] will differ from that of [tex]\( f(x) \)[/tex].
5. Horizontal asymptote:
- If [tex]\( f(x) \)[/tex] has a horizontal asymptote, this will be shifted vertically in both [tex]\( g(x) \)[/tex] and [tex]\( h(x) \)[/tex].
- For [tex]\( g(x) \)[/tex], any horizontal asymptote [tex]\( y = L \)[/tex] will be shifted to [tex]\( y = -L + 10 \)[/tex].
- For [tex]\( h(x) \)[/tex], any horizontal asymptote [tex]\( y = L \)[/tex] will be shifted to [tex]\( y = -L + 1 \)[/tex].
Thus, horizontal asymptotes of [tex]\( g(x) \)[/tex] and [tex]\( h(x) \)[/tex] will differ from those of [tex]\( f(x) \)[/tex].
6. x-intercept:
- For [tex]\( g(x) = -f(x) + 10 \)[/tex]:
- The x-intercept changes due to the reflection and upward shift. If [tex]\( f(a) = 0 \)[/tex], then [tex]\( g(a) = -f(a) + 10 = 10 \)[/tex]. New x-intercept needs to be recalculated where [tex]\( -f(x) + 10 = 0 \)[/tex].
- For [tex]\( h(x) = -f(x-4) + 1 \)[/tex]:
- The x-intercept changes due to the horizontal shift, reflection, and vertical shift. Originally, if [tex]\( f(a) = 0 \)[/tex], the new intercept would come from solving [tex]\( -f(b-4) + 1 = 0\)[/tex], which changes its position.
Therefore, x-intercepts of both [tex]\( g(x) \)[/tex] and [tex]\( h(x) \)[/tex] differ from [tex]\( f(x) \)[/tex].
To summarize, the features of the graphs of functions [tex]\( g \)[/tex] and [tex]\( h \)[/tex] that differ from those of [tex]\( f \)[/tex] are:
- Range
- y-intercept
- End behavior
- Horizontal asymptote
- x-intercept
1. Range:
- [tex]\( g(x) = -f(x) + 10 \)[/tex]:
- The "-f(x)" part reflects the graph of [tex]\( f(x) \)[/tex] over the x-axis, effectively flipping the range. Therefore, if [tex]\( f(x) \)[/tex] has values from [tex]\( a \)[/tex] to [tex]\( b \)[/tex], after reflection, those values become [tex]\( -a \)[/tex] to [tex]\( -b \)[/tex].
- The "+10" part shifts the entire graph up by 10 units. This means each value in the range will increase by 10 units. Therefore, the new range will be from [tex]\( -a + 10 \)[/tex] to [tex]\( -b + 10 \)[/tex].
- [tex]\( h(x) = -f(x-4) + 1 \)[/tex]:
- The "-f(x-4)" part first shifts the graph of [tex]\( f(x) \)[/tex] four units to the right and then reflects it over the x-axis, thus flipping the range.
- The "+1" part then shifts the graph up by 1 unit, resulting in an adjusted range from [tex]\( -a + 1 \)[/tex] to [tex]\( -b + 1 \)[/tex].
Thus, the range of both [tex]\( g(x) \)[/tex] and [tex]\( h(x) \)[/tex] will differ from [tex]\( f(x) \)[/tex].
2. Domain:
- Neither vertical reflections nor shifts (horizontal or vertical) affect the domain of the function. The domain of [tex]\( g(x) \)[/tex] and [tex]\( h(x) \)[/tex] remains the same as [tex]\( f(x) \)[/tex].
3. y-intercept:
- For [tex]\( g(x) = -f(x) + 10 \)[/tex]:
- The y-intercept of [tex]\( f(x) \)[/tex] is affected by both the reflection and the vertical shift. If [tex]\( f(0) = c \)[/tex], then the y-intercept of [tex]\( g(x) \)[/tex] will be [tex]\( -c + 10 \)[/tex].
- For [tex]\( h(x) = -f(x-4) + 1 \)[/tex]:
- The y-intercept of [tex]\( h(x) \)[/tex] is found by evaluating [tex]\( h(0) \)[/tex], which is [tex]\( -f(-4) + 1 \)[/tex]. This shifts the y-intercept to the point [tex]\( -f(-4) + 1 \)[/tex].
Therefore, the y-intercept for both [tex]\( g(x) \)[/tex] and [tex]\( h(x) \)[/tex] is different from that of [tex]\( f(x) \)[/tex].
4. End behavior:
- [tex]\( g(x) = -f(x) + 10 \)[/tex]:
- The end behavior of [tex]\( f(x) \)[/tex] is first reflected about the x-axis and then shifted up by 10 units.
- [tex]\( h(x) = -f(x-4) + 1 \)[/tex]:
- The end behavior of [tex]\( f(x) \)[/tex] is first shifted right by 4 units, reflected, and then shifted up by 1 unit.
The end behavior of both [tex]\( g(x) \)[/tex] and [tex]\( h(x) \)[/tex] will differ from that of [tex]\( f(x) \)[/tex].
5. Horizontal asymptote:
- If [tex]\( f(x) \)[/tex] has a horizontal asymptote, this will be shifted vertically in both [tex]\( g(x) \)[/tex] and [tex]\( h(x) \)[/tex].
- For [tex]\( g(x) \)[/tex], any horizontal asymptote [tex]\( y = L \)[/tex] will be shifted to [tex]\( y = -L + 10 \)[/tex].
- For [tex]\( h(x) \)[/tex], any horizontal asymptote [tex]\( y = L \)[/tex] will be shifted to [tex]\( y = -L + 1 \)[/tex].
Thus, horizontal asymptotes of [tex]\( g(x) \)[/tex] and [tex]\( h(x) \)[/tex] will differ from those of [tex]\( f(x) \)[/tex].
6. x-intercept:
- For [tex]\( g(x) = -f(x) + 10 \)[/tex]:
- The x-intercept changes due to the reflection and upward shift. If [tex]\( f(a) = 0 \)[/tex], then [tex]\( g(a) = -f(a) + 10 = 10 \)[/tex]. New x-intercept needs to be recalculated where [tex]\( -f(x) + 10 = 0 \)[/tex].
- For [tex]\( h(x) = -f(x-4) + 1 \)[/tex]:
- The x-intercept changes due to the horizontal shift, reflection, and vertical shift. Originally, if [tex]\( f(a) = 0 \)[/tex], the new intercept would come from solving [tex]\( -f(b-4) + 1 = 0\)[/tex], which changes its position.
Therefore, x-intercepts of both [tex]\( g(x) \)[/tex] and [tex]\( h(x) \)[/tex] differ from [tex]\( f(x) \)[/tex].
To summarize, the features of the graphs of functions [tex]\( g \)[/tex] and [tex]\( h \)[/tex] that differ from those of [tex]\( f \)[/tex] are:
- Range
- y-intercept
- End behavior
- Horizontal asymptote
- x-intercept
