State whether there is a possible transformation of f in the form g(x) = f(x-k). where k is a constant, for which the graph of y = g(x) will have no x-intercepts. Justify your answer.
Yes, there is a possible transformation of f in the form g(x) = f(x-k) for which the graph of y = g(x) will have no x-intercepts. This is because if the original function f(x) has x-intercepts, then shifting the graph horizontally by k units to the right (i.e., g(x) = f(x-k)) will also shift the x-intercepts to the right by the same amount, and may move them completely out of the domain of the new function g(x). However, this is not necessarily guaranteed and depends on the properties of the original function f(x).
Yes, there is a possible transformation of f in the form g(x) = f(x-k) for the graph of y = g(x) to have no x-intercepts.
To justify this claim, we need to consider the effect of the transformation g(x) = f(x-k) on the x-intercepts of the graph.
An x-intercept occurs when the value of y is equal to zero. In other words, to find the x-intercepts, we need to solve the equation g(x) = 0.
Substituting g(x) = f(x-k) into the equation, we get f(x-k) = 0.
Now, let's think about what happens when we shift the function f by an amount k units to the right. This means that any x-intercept of f(x) would be shifted to the right by k units.
If we assume that f(x) has at least one x-intercept, then shifting it to the right by k units would still result in an x-intercept, just at a different location. Therefore, g(x) = f(x-k) will always have x-intercepts if f(x) has at least one x-intercept.
On the other hand, if f(x) has no x-intercepts, then shifting it to the right by any amount k units will still not produce any x-intercepts in g(x). Therefore, g(x) = f(x-k) can have no x-intercepts if f(x) has no x-intercepts.
In conclusion, the possible transformation g(x) = f(x-k) can result in the graph of y = g(x) having no x-intercepts if and only if the function f(x) has no x-intercepts initially.