# Thread: Derivitives, finding lines

1. ## Derivitives, finding lines

The question, Find two lines that pass through (2 , 8) and are tangent to the curve y=x^3..how do you find slope?..

2. Originally Posted by jamman790
The question, Find two lines that pass through (2 , 8) and are tangent to the curve y=x^3..how do you find slope?..
To find the slope of the function, find the first derivative of $\displaystyle y=x^{3}$

Since you are asking, I am assuming that you don't fully understand derivatives. The derivative is the slope of the tangent line of a function. Since the slope of any function other than a line is constantly changing, you have to use derivatives to find the slope of the tangent line. To find the slope at that particular point, use the point they gave you (2,8).

3. Originally Posted by ahhh
To find the slope of the function, find the first derivative of $\displaystyle y=x^{3}$

Since you are asking, I am assuming that you don't fully understand derivatives. The derivative is the slope of the tangent line of a function. Since the slope of any function other than a line is constantly changing, you have to use derivatives to find the slope of the tangent line. To find the slope at that particular point, use the point they gave you (2,8).
I did that the derivitive is easy to find, becomes 3x^2 right? but there is two lines thus two possiblities, i dont think its asking for the equation of the tangent..but the equation of the line itself..so how do u find the slope? than how do you use that to find the two equations for the lines...?

4. Originally Posted by jamman790
I did that the derivitive is easy to find, becomes 3x^2 right? but there is two lines thus two possiblities, i dont think its asking for the equation of the tangent..but the equation of the line itself..so how do u find the slope? than how do you use that to find the two equations for the lines...?
Since the first derivative gives you the slope (we'll denote m=slope), then $\displaystyle m=3x^{2}$ right?

Use the equation $\displaystyle y=mx+b$ to find the exact tangent line.

Since they want the line passing through the point (2,8), x=2 and y= 8 right?

$\displaystyle 8=(3(2)^{2})(2)+b$ and $\displaystyle b=-16$ right?

So one of the lines is $\displaystyle y=(3x^{2})(x)-16$

And can be simplified to $\displaystyle y=3x^{3}-16$

5. sighh...lol sadly the teacher gave us two awnsers and they were like equations..thats the original awnser i got, but he said now...:P..blahh