Question 942469
It's easier to just complete the square but ...
The slope of the tangent is also the value of the derivative.
{{{y=x^2-4x+1}}}
{{{dy/dx=2x-4}}}
Set this equal to zero.
{{{2x-4=0}}}
{{{2x=4}}}
{{{x=2}}}
So then,
{{{y=a(x-2)^2+b}}}
{{{y=a(x^2-4x+4)+b}}}
{{{y=ax^2-4ax+4a+b}}}
Comparing to the original equation,
{{{-4a=-4}}}
{{{a=1}}}
Then,
{{{4a+b=1}}}
{{{4+b=1}}}
{{{b=-3}}}
So the equation in vertex form is,
{{{y=(x-2)^2-3}}}
and the vertex is (2,-3)
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{{{graph(300,300,-5,5,-5,5,x^2-4x+1)}}}