What is the equation of the line tangent to # f(x)=x(x-3)^2 # at # x=2 #?

Question

What is the equation of the line tangent to #  f(x)=x(x-3)^2 # at #  x=2  #?

Emilia Aguilar · Accepted Answer

#y=-3x+8#

Determine where the tangent line will cross:

#f(2)=2(2-3)^2=2(-1)^2=2#

The tangent line will pass through the point #(2,2)#.

To find the slope of the tangent line, find the value of the derivative at #x=2#.

To find #f'(x)#, we will use the product rule.

#f'(x)=(x-3)^2d/dx[x]+xd/dx[(x-3)^2]#

Find the value of each derivative. The second will require a simple application of the chain rule. Alternatively, you could say that #(x-3)^2=x^2-6x+9#, which makes the derivative findable through the power rule.

#f'(x)=(x-3)^2(1)+x(2(x-3))#

There is a simpler way to do this.

#f'(x)=x^2-6x+9+2x^2-6x#

#f'(x)=3x^2-12x+9#

Note that the derivative also could have been found through distribution of #f(x)#, as follows:

#f(x)=x(x^2-6x+9)#

#f(x)=x^3-6x^2+9x#

#f'(x)=3x^2-12x+9#

The derivative is the same no matter how you arrived at it. With the derivative in hand, we can calculate the tangent line's slope:

#f'(2)=3(2^2)-12(2)+9=12-24+9=-3#

The tangent line passes through the point #(2,2)# and has slope #-3#, giving the linear equation

#y-2=-3(x-2)#

or

#y=-3x+8#

The original function and its tangent line are graphed:

graph{(y+3x-8)=0 [-1, 5, -2.526, 6.364]}(x(x-3)^2-y)

Samantha Adkison · Answer

y + 3x - 8 = 0 The tangent's equation is y - b = m(x - a). where (a, b) is a line point and m= gradient. It is necessary to find m and (a, b). The tangent gradient is f'(x) and a = 2. Enter x = 2 into f(x) to find b. Differentiate using # color(blue)(" product and chain rules ")# f'(x) # = x d/dx(x-3)^2 + (x-3)^2 d/dx (x) # # = x [2(x-3) d/dx(x-3)] + (x-3)^2 .1# # = 2x(x-3) .1 + (x-3)^2 # = (x-3)(2x + x - 3 ) = 3 (x-3)(x-1) m = f'(2) = 3(2-3)(2-1) = -3 at this point and b = f(2) =# 2(-1)^2 = 2# The formula is y-2 = -3(x-2) y - 2 therefore equals -3x + 6. y + 3x - 8 = 0 then.

Isaiah Allen · Answer

undefined The equation of the line tangent to f(x)=x(x-3)^2 at x=2 is y = -4x + 16.