Check Work ?
Let f(x)=(x-6)(6x+4) Find the equation for the tangent line to the graph of f at x=1
Tangent line: y= ?
My work.....
f'(x)=(x-6)(6)+(1)(6x+4) I applied product rule
=12x-32
f'(1)=12(1)-32 = -20
y-(-50)=-20(x-1) [the -50 I got from plugging 1 into orig eq.]
y=-20x-30 ?

Question

Check Work ?
Let f(x)=(x-6)(6x+4) Find the equation for the tangent line to the graph of f at x=1 
Tangent line: y= ?
My work.....
f'(x)=(x-6)(6)+(1)(6x+4) I applied product rule
       =12x-32
f'(1)=12(1)-32 = -20
y-(-50)=-20(x-1) [the -50 I got from plugging 1 into orig eq.]
y=-20x-30 ?

whpalmer4 · Answer

Let's check by multiplying the expression for $f(x)$ and taking the derivative of that.
$$f(x) = (x-6)(6x+4) = 6x^2+4x-36x-24 = 6x^2-32x-24$$$$\frac{df(x)}{dx} =2*6x^{2-1}-1*32x^{1-1} = 12x-32$$So far so good.
$$f'(1) = 12(1)-32 = -20$$

$$f(1) = (1-6)(6(1)+4) = -5*10 = -50$$So we have to run our tangent line through the point $(1,-50)$
Using point-slope formula:$$y-y_0 = m(x-x_0)$$$$y-(-50) = -20(x-1)$$$$y+50 = -20x+20$$$$y=-20x-30$$

I agree with your answer.  Is it possible that you are supposed to provide it in different form?  100% certain you copied the problem correctly?

anonymous · Answer

This makes sense I must've misread the question or put the wrong answer. Thank you for helping me out :) and for the detailed explanation !

whpalmer4 · Answer

You're welcome!