How do you find the equation of the line tangent to #y= -x^3+6x^2-5x# at (1,0)?

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Answer 1 · 2016-10-21T16:38:00

#y=4x-4#

To find the tangent we will use the equation
#(y-y_1)=m(x-x_1)#,

where #(x_1,y_1)# is the point at which we are finding the tangent, and #m# is the gradient.

Since we know #(x_1,y_1)# we have to find the gradient of the tangent, this is done by differentiation.

#y=-x^3+6x^2-5x#

#(dy)/dx=-3x^2+12x-5#

#m=[(dy)/dx]_(x=1)=-3xx1^2+12xx1-5#

#m=[(dy)/dx]_(x=1)=-3+12-5=4#

eqn of tangent

#(y-0)=4(x-1)#

#=>y=4x-4#