Math
posted by Paul .
Find to find equation of the tangent line y=x^4x^3x^2x+1 at the point x=1

Is the calculus?
If so, take the derivative of y which would be 4x^3  3x^22x 1
Let x =1 to find the slope
4321 = m = 2
find the value of y when x = 1
You will then have the slope and a point to use to find the equation using y=mx+b 
so i found the y=1
then i subbed into y=mx+b and got 1=2(1)+b
1=b
then my equation would be 1=2(1)+1
is that right?
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