SOLUTION: Find the minimum slope of the line tangent to the curve y = x³-3x²+6x+3.

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Question 1111276: Find the minimum slope of the line tangent to the curve y = x³-3x²+6x+3.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The slope of the line tangent to the curve y+=+x%5E3-3x%5E2%2B6x%2B3
for each value of x is the value of the derivative,
dy%2Fdx=3x%5E2-6x%2B5 .


slope=3x%5E2-6x%2B5 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%5E2-6x%2B5 by looking at
zeros and changes of sign of d%28slope%29%2Fdx=6x-6 .
As d%28slope%29%2Fdx%3C0 for x%3C1 ,
d%28slope%29%2Fdx=0 for x=1 , and
d%28slope%29%2Fdx%3E0 for x%3E1 ,
there is a minimum at x=1 .
The minimum slope, for x=1 is
slope=3%2A1%5E2-6%2A1%2B5=3-6%2B5=2 .

Alternately, if you memorized class information,
you know that the expression ax%5E2%2Bbx%2Bc with a%3E0
has a minimum at x=%28-b%29%2F%222+a%22 .
Maybe your teacher expects you to do that,
find that the minimum is at x=%28-%28-6%29%29%2F%282%2A3%29=6%2F6=1 ,
and calculate that for x=1 ,
slope=3%2A1%5E2-6%2A1%2B5=3-6%2B5=2 .

You could just use your algebra knowledge (no memorization)
and "complete the square" like this
slope=3x%5E2-6x%2B5
slope%2F3=x%5E2-2x%2B5%2F3
slope%2F3=%28x-1%29%5E2%2B5%2F3-1
slope%2F3=%28x-1%29%5E2%2B2%2F3
slope=3%28%28x-1%29%5E2%2B2%2F3%29
slope=3%28x-1%29%5E2%2B2%29
As %28x-1%29%5E2%3E=0 , slope%3E=3