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Math Help - The slope of a tangent

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    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.
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  2. #2
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    Quote Originally Posted by darksoulzero View Post
    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?
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  3. #3
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    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.
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    Quote Originally Posted by darksoulzero View Post
    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 )
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    Quote Originally Posted by Also sprach Zarathustra View Post
    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?
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    Quote Originally Posted by darksoulzero View Post
    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.
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    Quote Originally Posted by HallsofIvy View Post
    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.
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    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.
    Last edited by HallsofIvy; July 11th 2010 at 07:21 AM.
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