Question 1111276
The slope of the line tangent to the curve {{{y = x^3-3x^2+6x+3}}}
for each value of {{{x}}} is the value of the derivative,
{{{dy/dx=3x^2-6x+5}}} .
{{{drawing(300,300,-4,6,-20,20,
graph(300,300,-4,6,-20,20,x^3-3x^2+6x+3,3.75x+3.5,6x-1),
green(circle(0.5,5.375,0.1)),blue(circle(2,11,0.1)),
locate(0.9,5,blue(slope=6)),locate(0.8,3,blue(at)),locate(1.5,3,blue(x=2)),
locate(3,15.3,green(slope=3.75)),locate(3,13.3,green(at)),locate(3.7,13.3,green(x=0.5))
)}}}


{{{slope=3x^2-6x+5}}} is a quadratic polynomial and graphs as a parabola.
You are asked for the minimum value of that expression,
and there are several ways to get to the result.
Can you figure out what way your teacher would favor?
 
Maybe your teacher expects you to find the minimum of
{{{slope=3x^2-6x+5}}} by looking at
zeros and changes of sign of {{{d(slope)/dx=6x-6}}} .
As {{{d(slope)/dx<0}}} for {{{x<1}}} ,
{{{d(slope)/dx=0}}} for {{{x=1}}} , and
{{{d(slope)/dx>0}}} for {{{x>1}}} ,
there is a minimum at {{{x=1}}} .
The minimum slope, for {{{x=1}}} is
{{{slope=3*1^2-6*1+5=3-6+5=2}}} .
 
Alternately, if you memorized class information,
you know that the expression {{{ax^2+bx+c}}} with {{{a>0}}}
has a minimum at {{{x=(-b)/"2 a"}}} .
Maybe your teacher expects you to do that,
find that the minimum is at {{{x=(-(-6))/(2*3)=6/6=1}}} ,
and calculate that for {{{x=1}}} ,
{{{slope=3*1^2-6*1+5=3-6+5=2}}} .
 
You could just use your algebra knowledge (no memorization)
and "complete the square" like this
{{{slope=3x^2-6x+5}}}
{{{slope/3=x^2-2x+5/3}}}
{{{slope/3=(x-1)^2+5/3-1}}}
{{{slope/3=(x-1)^2+2/3}}}
{{{slope=3((x-1)^2+2/3)}}}
{{{slope=3(x-1)^2+2)}}}
As {{{(x-1)^2>=0}}} , {{{slope>=3}}}