Question 458808
I'm 46 years old and am trying to relearn algebra on my own.  Any help is greatly appreciated.

For the function f(x) = x² - 2x + 1
           (a) find f(0)
<pre>
Notice that the difference between f(x) and f(0) is that
f(0) has a (0) and f(x) has an (x).  
So everywhere there is an x in the right side,
put a (0) there instead.  In other words instead of

f(x) = x² - 2x + 1 

you will have

f(0) = (0)² - 2(0) + 1

Now (0)² is just 0 and 2(0) is also just 0 so the above becomes

f(0) = 0 - 0 + 1

and that becomes just

f(0) = 1 

That's the answer!

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(b) solve f(x) = 0

Instead of f(x) write the right side of the given equation
for f(x), which is x² - 2x + 1, so instead of

       f(x) = 0

You will have

x² - 2x + 1 = 0

Now you have to factor the left side and get

(x - 1)(x - 1) = 0

Now you put = 0 after each of those factors:

x - 1 = 0           x - 1 = 0

Then you add 1 to both sides to get rid of the -1
on the left and get

    x = 1               x = 1

So you get the same answer twice.  So there is
just one solution, namely

 x = 1

Edwin</pre>