What is the equation of the line tangent to # f(x)=(3x-1)(2x+4) # at # x=-1 #?

Question

What is the equation of the line tangent to #  f(x)=(3x-1)(2x+4)  # at #  x=-1  #?

Christopher Allers · Accepted Answer

#y=-2x-10#

Find #f'(-1)# for the slope of the tangent line at the point #(-1,-8)#.

#f(x)=6x^2+10x-4#

Recall that #d/dx[x^n]=nx^(n-1)#.

#f'(x)=12x+10#

#f'(-1)=-2#

We can express the tangent line's equation in point-slope form as follows:

#y+8=-2(x+1)#

Forming the slope-intercept:

#y=-2x-10#

Elijah Abram · Answer

undefined To find the equation of the line tangent to the function f(x)=(3x-1)(2x+4) at x=-1, we need to find the derivative of the function and evaluate it at x=-1.

First, let's find the derivative of f(x):
f'(x) = (3x-1)(d/dx)(2x+4) + (2x+4)(d/dx)(3x-1)
       = (3x-1)(2) + (2x+4)(3)
       = 6x - 2 + 6x + 12
       = 12x + 10

Next, let's evaluate the derivative at x=-1:
f'(-1) = 12(-1) + 10
       = -12 + 10
       = -2

Now, we have the slope of the tangent line, which is -2. To find the equation of the line, we also need a point on the line. Since the line is tangent to the function at x=-1, we can use this point to find the y-intercept.

Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope, we can substitute the values:
y - f(-1) = -2(x - (-1))
y - f(-1) = -2(x + 1)
y - f(-1) = -2x - 2
y = -2x - 2 + f(-1)

Finally, we substitute f(-1) into the equation:
y = -2x - 2 + f(-1)
y = -2x - 2 + (3(-1)-1)(2(-1)+4)
y = -2x - 2 + (-4)(2+4)
y = -2x - 2 + (-4)(6)
y = -2x - 2 - 24
y = -2x - 26

Therefore, the equation of the line tangent to f(x)=(3x-1)(2x+4) at x=-1 is y = -2x - 26.