# Thread: The slope of a tangent

1. ## The slope of a tangent

I am trying to do this question with out derivatives for about 45 min and i still don't know what to do. any help would be appreaciated. Thank you.

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

I know that the slope of the tangent is 8, but i am not sure how to proceed.

2. Originally Posted by darksoulzero
I am trying to do this question with out derivatives for about 45 min and i still don't know what to do. any help would be appreaciated. Thank you.

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

I know that the slope of the tangent is 8, but i am not sure how to proceed.
Why are you trying to do it without using calculus?

3. its a thinking question in my textbook they teach the slope of the tanget line and how to determine it by using the difference quoient. I didn't learn derivative yet so i'm trying to see how to determine it.

4. Originally Posted by darksoulzero
its a thinking question in my textbook they teach the slope of the tanget line and how to determine it by using the difference quoient. I didn't learn derivative yet so i'm trying to see how to determine it.
You later study that is the same thing(difference quotient & derivative )

5. Originally Posted by Also sprach Zarathustra
You later study that is the same thing(difference quotient & derivative )
ah ok. So I know that the slop of the tangent line is parallel to the line y=8x so that means that the difference quotient of the function 3x^2-4x would give me 8. how would I deterine the coordinates with this information?

6. Originally Posted by darksoulzero
ah ok. So I know that the slop of the tangent line is parallel to the line y=8x so that means that the difference quotient of the function 3x^2-4x would give me 8. how would I deterine the coordinates with this information?
Well, first, the "slope" is a number and so is not "parallel" to anything, but I understand what you mean- the tangent line is parallel to y= 8x and so the slope is 8. The limit of the difference quotient must be 8. The difference quotient is $\frac{3(x+h)^2- 4(x+h)- 3x^2- 8}{h}$. Taking the limit as h goes to 0 gives you a function of x. Set it equal to 8 and solve for x.

7. Originally Posted by HallsofIvy
Well, first, the "slope" is a number and so is not "parallel" to anything, but I understand what you mean- the tangent line is parallel to y= 8x and so the slope is 8. The limit of the difference quotient must be 8. The difference quotient is $\frac{3(x+h)^2- 4(x+h)- 3x^2- 8}{h}$. Taking the limit as h goes to 0 gives you a function of x. Set it equal to 8 and solve for x.
Thank you for the reply, but i figured out how to deal with the question. Apparently in my frustration i just messed up my algebra.

8. By the way, another way to find a tangent is to use Fermat's method of "ad-equation" which he developed literally "precalculus". Obviously, $3x^2- 4x$ will be equal to $y= ax+ b$ where they intersect- but there will be a double root for (x, y) where they are tangent. The quadratic equation $3x^2- (4- a)x- b= 0$ will have a double root where its discriminant, $(4- a)^2+ 12b= 0$. Knowing that the slope is a= 8 reduces that to a simple equation for b and then you can find the correct x and y values.