Results 1 to 6 of 6

Math Help - Finding the equations of tangents

  1. #1
    Junior Member
    Joined
    Feb 2011
    Posts
    72

    Finding the equations of tangents

    Finding the equations of the tangent to the curve at the given point.

    y = 4/x^2 at x = 1

    I understand how to work out this but I haven't figured out how to treat fractions. For example I can workout y = 5x^2 + 3.

    Please help as this for a test and I need to figure out how to work it out.

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Recall your laws of exponents

    \dfrac{4}{x^2} = 4x^{-2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2011
    Posts
    72
    Quote Originally Posted by e^(i*pi) View Post
    Recall your laws of exponents

    \dfrac{4}{x^2} = 4x^{-2}
    Thanks.

    But how can I work out:

    4(2+h)^2

    I have tried working it work but I couldn't. I am trying to find the y axis value to replace (x, ?)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    use the alternate definition of a derivative ...

    \displaystyle f'(a) = \lim_{x \to a} \dfrac{f(x) - f(a)}{x-a}

    ... where a = 1
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,306
    Thanks
    1282
    Quote Originally Posted by Googl View Post
    Thanks.

    But how can I work out:

    4(2+h)^2

    I have tried working it work but I couldn't. I am trying to find the y axis value to replace (x, ?)
    You did not say that you have to find the derivative from the "difference quotient" formula.

    I assume you mean 4(2+ h)^{-2}. To do that, I recommend you go back to the fraction form: \frac{4}{x^2} so that the difference quotient is \frac{\frac{4}{(2+h)^2}- \frac{4}{2^2}}{h}.

    To simplify that, get "common denominators". \frac{4}{(2+h)^2}- 1= \frac{4}{(2+h)^2}- \frac{(2+h)^2}{(2+h)<br />
^2} = \frac{4- (4+ 4h+ h^2)}{(2+h)^2}= \frac{-4h- h^2}{(2+ h)^2}

    Dividing that by h, \frac{-4h- h^2}{h(2+ h)^2}= \frac{h(-4- h)}{h(2+h)^2}= \frac{-4-h}{(2+ h)^2}

    Now, take the limit of that as h goes to 0.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Feb 2011
    Posts
    72
    Hello HallsofIvy,

    Your working out is not what I am used to. How you have managed to transform the fractions... And I have to learn how to transform it the way you have. I understand how to find the equations of the tangent to the curve at the given point. This is what I have been revising at the moment, but when it came to the y equalling to a fraction curve that's where I got stuck. My reference doesn't actually show how to work out this. It shows for the standard ones without fractions but in the practice question it has some of the questions as fractions.

    I understand I am supposed to use y1 - y = m(x - x1) where m is the gradient. So I am supposed to find the gradient first which I know how to, and then y1 and x1 and replace them in the equation. That's how I am working it out at the moment.

    In your reply I don't understand why you had to subtract by 1. And why you concentrated on just the denominator towards the end, also I didn't quite understand how the final step could be useful.

    I think the best way for me to figure out how to handle these types of questions is to get the full working out for the y = 4/x^2 at x = 1 (finding the equations of the tangent to the curve at the given point.)

    Thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Equations of all tangents..
    Posted in the Calculus Forum
    Replies: 12
    Last Post: November 23rd 2010, 05:07 PM
  2. Deduce equations of tangents
    Posted in the Geometry Forum
    Replies: 6
    Last Post: May 30th 2010, 11:06 PM
  3. Replies: 7
    Last Post: July 4th 2009, 10:48 PM
  4. Circles- tangents and equations
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: March 7th 2009, 05:09 AM
  5. Tangents - Parametric Equations
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 26th 2008, 03:55 PM

Search Tags


/mathhelpforum @mathhelpforum