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Question 1196152: Hi, I have no idea how to prove this. Any help appreciated. Thank you
Let f(x) = ax^3 + bx^2 + cx + d and suppose that r is a root of the equation f(x) = 0.
(a) Show that r − h is a root of f(x + h) = 0.
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52810)   (Show Source):
You can put this solution on YOUR website! .
Hi, I have no idea how to prove this. Any help appreciated. Thank you
Let f(x) = ax^3 + bx^2 + cx + d and suppose that r is a root of the equation f(x) = 0.
(a) Show that r − h is a root of f(x + h) = 0.
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Substitute x= r-h into f(x+h).
You will get f(x+h) = f((r-h)+h) = f(r) = 0, since "r" is the root of f(x), as it is given in the problem.
It PROVES that r-h is the root of f(x+h) = 0.
QED, which means " the proof is complete ".
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
We don't have to restrict f(x) to be of the form ax^3 + bx^2 + cx + d. The function f(x) can be anything really.
All that matters is that r is a root of f(x), i.e. f(r) = 0
The input x = r leads to the output f(x) = 0
We say that r is a zero of f(x), which in my opinion can be misleading because r itself may be zero or nonzero.
So if we wanted to see if x = r-h is a root, then,
f(x+h) = f( (r-h)+h )
f(x+h) = f( (r-h)+h )
f(x+h) = f( r + (-h+h) )
f(x+h) = f( r + 0 )
f(x+h) = f( r )
f(x+h) = 0
Therefore, if f(r) = 0, then r-h has been confirmed to be a root of f(x+h) = 0
I.e. if r is a root of f(x), then r-h is a root of f(x+h).
What's really happening is the move from x to x+h shifts the xy axis h units to the right.
This gives the illusion the f(x) curve shifts h units to the left.
That's why the root x = r shifts h units to the left to get to x = r-h
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An example:
we have x = 5 as a root because
So r = 5 in this case
Now consider shifting the xy axis h = 3 units to the right
The old input x is now x+h, in this case x+3
Shifting the xy axis 3 units to the right gives the illusion the exponential curve shifts 3 units to the left.
So we'll be shifting the old root (r = 5) to its new spot of r-h = 5-3 = 2 which is also a shift of 3 units to the left.
Then notice the following:
 Plug in h = 3
 Replace every x with r-h
 Plug in r = 5 and h = 3
We arrive back at the same conclusion as before, except this time we have a specific numeric example rather than a more vague algebraic statement.
Of course one numeric example is not enough to prove a theorem entirely (we would have to check the infinitely many numbers on the number line which isn't feasible). It's sole purpose is to help illustrate what's going on. I recommend graphing out your favorite function to get practice with this. Desmos is a good free tool to do such a thing.
Once again it could be of the form ax^3 + bx^2 + cx + d, but it doesn't have to be.
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