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The transformation that maps the graph of `y = sqrt(8x^3 + 1)` onto the graph of `y = sqrt(x^3 + 1)` is a
`A`
| `text(Let)\ f(x)` | `= sqrt(8x^3 + 1)` |
| `f(1/2 x)` | `= sqrt(8(1/2 x)^3 + 1)` |
| `= sqrt(x^3 + 1)` |
`:.\ text(Transformation correct when)\ \ x\ \ text(is swapped for)\ \ x/2`
`text(i.e. graph is dilated by factor of 2 from)\ ytext(-axis)`
`=> A`
The point `A (3, 2)` lies on the graph of the function `f(x)`. A transformation maps the graph of `f(x)` to the graph of `g(x)`,
where `g(x) = 1/2 f(x - 1)`. The same transformation maps the point `A` to the point `P`.
The coordinates of the point `P` are
A. `(2, 1)`
B. `(2, 4)`
C. `(4, 1)`
D. `(4, 2)`
`C`
`g(x) = 1/2 f(x – 1),\ A(3, 2)`
`text(Dilate by a factor of)\ 1/2\ text(from)\ x text(-axis:)`
`A(3, 2) -> A′(3, 1)`
`text(Translate 1 unit to right:)`
`A′(3, 1) -> P(4, 1)`
`=> C`
The graph of a function `f(x)` is obtained from the graph of the function `g(x) = sqrt (2x - 5)` by a reflection in the `x`-axis followed by a dilation from the `y`-axis by a factor of `1/2`.
Which one of the following is the function `f(x)`?
A. `f(x) = sqrt (5 - 4x)`
B. `f(x) = - sqrt (x - 5)`
C. `f(x) = sqrt (x + 5)`
D. `f(x) = −sqrt (4x - 5)`
`D`
`text(Let)\ \ y=sqrt(2x-5)`
`text(1st transformation:)`
`y = – sqrt(2x-5)`
COMMENT: Using “swap” terminology for dilations from the y-axis is simpler and more intelligible for students in our view.
`text(2nd transformation:)`
`text(Dilate from)\ y text(-axis by a factor of)\ 1/2`
`=>\ text(Swap)\ \ x → 2x`
| `y` | `=-sqrt(2(2x)-5)` |
| `=- sqrt(4x-5)` | |
| `:. f(x)` | `= −sqrt(4x – 5)` |
`=> D`
The point `P\ text{(4, −3)}` lies on the graph of a function `f(x)`. The graph of `f(x)` is translated four units vertically up and then reflected in the `y`-axis.
The coordinates of the final image of `P` are
`A`
`text(1st transformation:)`
`P(4,−3)\ ->\ (4,1)`
`text(2nd transformation:)`
`(4,1)\ ->\ (-4,1)`
`=> A`
The graph of the function `f(x) = 3x^(5/2)` is reflected in the `x`-axis and then translated 3 units to the right and 4 units down.
The equation of the new graph is
A. `y = 3(x - 3)^(5/2) + 4`
B. `y = -3 (x - 3)^(5/2) - 4`
C. `y = -3 (x + 3)^(5/2) - 1`
D. `y = -3 (x - 4)^(5/2) + 3`
`B`
`text(Let)\ \ y= 3x^(5/2)`
`text(Reflect in the)\ x text(-axis:)`
`y= – 3x^(5/2)`
`text(Translate 3 units to the right:)`
`y=- 3(x-3)^(5/2)`
`text(Translate 4 units down:)`
`y=- 3(x-3)^(5/2) – 4`
`=> B`