Question 424266
For the function f(x)=x^2-2x+1, 
(a)find f(1)

<pre><font face = "consolas" color = "indigo" size = 4><b>

f(<font color="red">x</font>) = <font color="red">x</font>² - 2<font color="red">x</font> + 1

Replace every <font color="red">x</font>
 with <font color="red">(1)</font>


f(<font color="red">(1)</font>) = <font color="red">(1)</font>² - 2<font color="red">(1)</font> + 1

We don't need the double parentheses after the f, and we will
drop the color scheme:

f(1) = (1)² - 2(1) + 1

Now we simplify the right side:

f(1) = 1 - 2 + 1

f(1) = 0
</pre></font></b>
(b) solve f(x)=0 
<pre><font face = "consolas" color = "indigo" size = 4><b>

f(x) = x² - 2x + 1

Put a zero where the f(x) is

  0 = x² - 2x + 1

Swap sides:

 x² - 2x + 1 = 0

Factor the left side:

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

Use the zero-factor principle:

x - 1 = 0         x - 1 = 0
    x = 1             x = 1

</pre></font></b>
Graph the function y=f(x)=x^2+1
<pre><font face = "consolas" color = "indigo" size = 4><b>

Get some points:

 x | y =  f(x) =    x² + 1             | point
-------------------------------------------------
-2 | y = f(-2) = (-2)² + 1 = 4 + 1 = 5 | (-2,5) 
-1 | y = f(-1) = (-1)² + 1 = 1 + 1 = 2 | (-1,2)
 0 | y =  f(0) =  (0)² + 1 = 0 + 1 = 1 |  (0,1}
 1 | y =  f(1) =  (1)² + 1 = 1 + 1 = 2 |  (1,2) 
 2 | y =  f(2) =  (2)² + 1 = 4 + 1 = 5 |  (2,5)

Plot those points:

{{{drawing(2400/7,400,-3,3,-1,6, graph(2400/7,400,-3,3,-1,6),

circle(-2,5,.05), locate(-2,5,"(-2,5)"),
circle(-1,2,.05), locate(-1,2,"(-1,2)"),

circle(0,1,.05), locate(0,1,"(0,1)"),

circle(1,2,.05), locate(1,2,"(1,2)"),

circle(2,5,.05), locate(2,5,"(2,5)") )}}}

Now draw a smooth U-shaped curve through those points:

{{{drawing(2400/7,400,-3,3,-1,6, graph(2400/7,400,-3,3,-1,6,x^2+1),

circle(-2,5,.05), locate(-2,5,"(-2,5)"),
circle(-1,2,.05), locate(-1,2,"(-1,2)"),

circle(0,1,.05), locate(0,1,"(0,1)"),

circle(1,2,.05), locate(1,2,"(1,2)"),

circle(2,5,.05), locate(2,5,"(2,5)") )}}}

Edwin</pre></font></b>