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

Question

Christopher Allers · Accepted Answer

I would find the derivative of your function, #y'(x)#, evaluate the derivative at the point of interest (#x=2#) and use it as the slope of the line through #(2,-2)#.
Step 1
Derive: #y'(x)=3x^2-6x+1#
Step2:
Evaluate: #y'(2)=12-12+1=1#
Step3:
Tangent: line through #(2,-2)# and slope #1#;
#y-(-2)=1*(x-2)#
#y+2=x-2#
#y=x-4# which is your tangent.

Graphically:

Samantha Adkison · Answer

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

First, we find the derivative of the function y=x^3-3x^2+x. The derivative is given by dy/dx = 3x^2 - 6x + 1.

Next, we substitute x=2 into the derivative to find the slope at the point (2,-2). Substituting x=2 into the derivative, we get dy/dx = 3(2)^2 - 6(2) + 1 = 12 - 12 + 1 = 1.

Therefore, the slope of the tangent line at the point (2,-2) is 1.

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 (2,-2) and m=1 into the equation.

Thus, the equation of the tangent line to the graph of y=x^3-3x^2+x at the point (2,-2) is y - (-2) = 1(x - 2), which simplifies to y + 2 = x - 2.