Answer:
The y-intercept of the function and the vertex of the function would change. (III and IV) This answer doesn't seem to be in the answer list (but there is proof for the answers below), though.
Step-by-step explanation:
(Notice that [tex]2f(x-4) = 2(x-4)^2.[/tex])
The domain of the function won't change because quadratics can all take all numbers as their input (which, by definition, is the domain of a function).
The range of the function won't change either, because we have just moved the function with 4 to the right (by substituting [tex]x[/tex] with [tex]x-4[/tex]) and multiplied the graph with respect to the y-axis by 2. The range will stay as all non-negative real numbers.
The y-intercept of the function will change, because for [tex]f(x)=x^2[/tex] it is [tex]y=0[/tex], while for [tex]y=2(x-4)^2[/tex] substituting [tex]x=0[/tex] doesn't give [tex]y=0[/tex]. (Instead, it gives [tex]y=32[/tex].)
The vertex of the function will change, because we have moved the function with 4 to the right and multiplied the graph with respect to the y-axis by 2: the vertex will have at the very least moved to the right.
