What is StartFraction 5 Over 8 EndFraction divided by One-fourth?
2 answers:
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Functions A and C and D are translations - this means that they are exactly the same shape as f(x), but are displaced on the axes. A translation in the y-direction by k units is represented by f(x) + k, and a translation in the x-direction by k units is represented by f(x - k). This can be summarised in column vector notation:
A translation of the graph y = f(x), by the vector![\left[\begin{array}{ccc}a\\b\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%5C%5Cb%5Cend%7Barray%7D%5Cright%5D%20)
results in the graph defined by y - b = f(x - a). This is done by replacing x with (x - a) and y with (y - b). Finally it is rearranged to make y the subject, as is often the convention: y = f(x - a) + b.
Since the graph needs to change shape, this requires a 'stretch'. A stretch is performed by multiplying x or y by a constant, which here is k.
To stretch in the x-direction by a factor n, replace x with (1/n)x. Similarly to stretch in the y-direction by a factor n, replace y with (1/n)y.
For example to stretch y = f(x) by factor n in the x-direction, it would become y = f((1/n)x).
A simple quadratic graph like y = x^2 behaves slightly differently to most other graphs, in that it can narrowed by a stretch in the x or y direction. This means that a more convenient option is to perform a stretch in the y direction, since the multiplier lies outside the function notation. The stretching factor is 3, so we replace y with (1/3)y:
(1/3)y = f(x)
y = 3f(x)
Since k = 3, this becomes
y = kf(x)
... and hence the correct answer is B
A translation of the graph y = f(x), by the vector
Since the graph needs to change shape, this requires a 'stretch'. A stretch is performed by multiplying x or y by a constant, which here is k.
To stretch in the x-direction by a factor n, replace x with (1/n)x. Similarly to stretch in the y-direction by a factor n, replace y with (1/n)y.
For example to stretch y = f(x) by factor n in the x-direction, it would become y = f((1/n)x).
A simple quadratic graph like y = x^2 behaves slightly differently to most other graphs, in that it can narrowed by a stretch in the x or y direction. This means that a more convenient option is to perform a stretch in the y direction, since the multiplier lies outside the function notation. The stretching factor is 3, so we replace y with (1/3)y:
(1/3)y = f(x)
y = 3f(x)
Since k = 3, this becomes
y = kf(x)
... and hence the correct answer is B