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

1 Answer
Jan 30, 2017

#y=3x-1#

Explanation:

The equation of the tangent in #color(blue)"point-slope form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y-y_1=m(x-x_1))color(white)(2/2)|)))#
where m represents the slope and # (x_1,y_1)" a point on the tangent"#

#color(orange)"Reminder: "m_("tangent")=dy/dx" at x=a"#

#y=3x^2-x^3#

#rArrdy/dx=6x-3x^2#

#"at "x=1tody/dx=6-3=3#

#"using "m=3" and "(x_1,y_1)=(1,2)" then"#

#y-2=3(x-1)larrcolor(red)"in point-slope form"#

distributing and simplifying gives an alternative version of the equation.

#y-2=3x-3#

#rArry=3x-1larrcolor(red)" in slope-intercept form"#