Find Point on Curve f(x) Where Tangent is Parallel to y=8x

[msimard8]

Find the coordinates of the point on the curve f(x)=3x^2-4x, where the tangent is parallel to the line y=8x.


Ok i know the slope of the tangent is 8.

and i know the formula is m=[f(a+h)-f(a)]/h

i need a hint please

 

[courtrigrad]

What is the derivative of $$f(x)$$? Use the power rule, set the derivative equal to 8, and solve for x. Plug x back into f(x), and you got the point.

 

[msimard8]

umm we havnt been taught derivative's yet. This is an introductory to it.

 

[courtrigrad]

your title is "limit question." have you been taught:

$$f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$$? Use this definition for $$f(x) = 3x^{2} - 4x$$.

 

[msimard8]

yes i have been taught that

but don't you need an x or y value of the tangent to use that formula.

 

[courtrigrad]

You know that the slope of the curve at the unknown point is 8 (because it is parallel to y = 8x). You don't need any values to use that formula. Just substitute (x+h) into f(x), and expand it out. After doing this subtract 3x^2 - 4x from f(x+h). Divide by h. Then substitute h = 0 to get f'(x).

 

[msimard8]

thanks
question solved