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

Question

Aurora Alvarado · Accepted Answer

#y=3x-1#

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"#

Axel Alvarez · Answer

undefined To find the equation of the line tangent to the curve y=3x^2-x^3 at the point (1,2), we need to find the slope of the tangent line at that point.

To find the slope, we take the derivative of the given function y=3x^2-x^3 with respect to x.

The derivative of y=3x^2-x^3 is dy/dx = 6x - 3x^2.

Now, substitute x=1 into the derivative to find the slope at the point (1,2).

dy/dx = 6(1) - 3(1)^2 = 6 - 3 = 3.

So, the slope of the tangent line at the point (1,2) is 3.

Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we can substitute the values to find the equation of the tangent line.

Using (1,2) as the point and 3 as the slope, we have y - 2 = 3(x - 1).

Simplifying the equation, we get y - 2 = 3x - 3.

Rearranging the equation, we have y = 3x - 1.

Therefore, the equation of the line tangent to y=3x^2-x^3 at the point (1,2) is y = 3x - 1.