If the graph of the function moves to the right two units what is the new vertex of the function

Let's talk through this a one step at a time.

*Since f(x) is concave-up with its vertex on the x-axis, we know f(x) ≥ 0.

*We also know that when we shift a function's domain by a positive number, we shift the function left and when we shift a function's domain by a negative number, we shift the function right. So f(x-5) is f(x) shifted to the right by 5.

*At this point, f(x-5) has its vertex at (5,0).

*When we negate f(x-5), the parabola becomes concave down yet the vertex remains at (5,0). Now we're at -f(x-5). At this point we have -f(x-5)≤0 with a range (-∞,0]

*If we add 2 to create g(x)=2-f(x-5), then we have a concave down parabola with its vertex shifted up by 2, at (5,2). So, g(x) is concave down with its vertex at (5,2). Hence

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v