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
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"#