Let f(x)=(x-6)(4x+2) find an equation for the tangent line to the graph of x=12?

The equation of a tangent line at a specific point with respect to a given function is called the derivative.

We start with:
#f(x) = (x-6)(4x+2)#
#f(x) = 4x^2-24x+2x-12#
#f(x) = 4x^2 -22x-12#

Taking the derivative:
#f'(x) = 8x - 22#

Assessing at #x=12#:
#f'(12) = 8(12)-22 = 74#

This means the slope of the tangent line is 74.

Now we plug in #x=12# into the original #f(x)#:
#f(12) = 4(12)^2 - 22(12)-12 = 300#

This means the tangent line touches #f(x)# at the point #(12,300)#.

The equation of a line is:
#(y - y_0) = m (x-x_0)#
#(y - 300) = 74 (x - 12)#
#y - 300 = 74x - 960#
#y = 74x - 660#

Hence, the tangent line equation is:
#y = 74x - 660#

Remember the product rule:

If #f(x)=g(x)*h(x)#, then #f'(x)=g'(x)*h(x)+g(x)*h'(x)#

Therefore,

#d/dx[(x-6)(4x+2)]=d/dx(x-6)*(4x+2)+(x-6)*d/dx(4x+2)#

#=>d/dx(x-6)*(4x+2)+(x-6)*d/dx(4x+2)#

We use the power rule:

#x^n=nx^(n-1)# where #n# is a constant.

When you are adding/subtracting a constant from a variable, then you could just eliminate it.

#=>1*x^(1-1)*(4x+2)+(x-6)*4*1x^(1-1)#

#=>(4x+2)+(x-6)*4#

#=>(4x+2)+(4x-24)#

#=>8x-22#

The slope of the tangent line is:

#f'(12)=8*12-22#

#f'(12)=74#

We also find #f(x)#

#f(12)=(12-6)(12*4+2)#

#f(12)=(6)(50)#

#f(12)=300#

We now use the slope-intercept form: #m(x-x_1)=y-y_1#

#74(x-12)=y-300#

#74x-888=y-300#

#74x-588=y#