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Math Help - Find the equation of the tangent line to y=1/2x^2 - 3 that has a slope of 4.?

  1. #1
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    Wink Find the equation of the tangent line to y=1/2x^2 - 3 that has a slope of 4.?

    So the two equations are
    1) Y = x^2 + 1
    2) Y = mx – 4
    Equate both the equations
    x^2+1 = mx -4
    x^2 +5 + mx = 0
    My teacher told me that from here use the quadratic formula in questions similar to these but in those questions the slope was given and the y-intercept wasn’t. so it worked out.. How do I find the value of m in the equations…
    Last edited by ruscutie100; October 18th 2007 at 10:20 PM. Reason: wrong
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  2. #2
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    Quote Originally Posted by ruscutie100 View Post
    So the two equations are
    1) Y = x^2 + 1
    2) Y = mx – 4
    Equate both the equations
    x^2+1 = mx -4
    x^2 +5 + mx = 0
    My teacher told me that from here use the quadratic formula in questions similar to these but in those questions the slope was given and the y-intercept wasn’t. so it worked out.. How do I find the value of m in the equations…
    Please ignore the topic..i edited it l8r on.. the question is
    Find the equations of the tangent lines to Y = X^2 +1 that have a y intercept of -4
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  3. #3
    Super Member angel.white's Avatar
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    Quote Originally Posted by ruscutie100 View Post
    Draw the parabola.
    Find the point of intersection.

    So i was trying to do this question. My teacher told me that whenever you see f(x) = 1/x then assume that it's the reciprocal function and the graph always takes the same shape...here'z the link for the shape:-http://www.gomath.com/htdocs/images/7-10...

    so shouldn't the equation y=1/2x^2 - 3 take the same shape... isn't it the reciprocal function because when i graph it on a graphing calculator it takes the shape of a parabola and i checked the answer the teacher gave us, and he also made it in a shape of parabola... can some1 explain me how do i come to know when is it reciprocal function and when is it a parabola
    Plz help me asap.. i have a test on this tomorrow... thnx

    (i don't know calculus)
    Having a hard time figuring out your formula
    y=1/2x^2 - 3

    does this equal
    a) y= \frac{1}{2}x^{2}-3

    b) y= \frac{1}{2x^{2}}-3

    c) y= \frac{1}{2x^{2}-3}
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  4. #4
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    Quote Originally Posted by angel.white View Post
    Having a hard time figuring out your formula
    y=1/2x^2 - 3

    does this equal
    a) y= \frac{1}{2}x^{2}-3

    b) y= \frac{1}{2x^{2}}-3

    c) y= \frac{1}{2x^{2}-3}
    srry i changed the questions... now i got it.. plz answer the edited one.. .thnx 4 replying 4 this question
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  5. #5
    Super Member angel.white's Avatar
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    Y = X^{2} +1 that have a y intercept of -4



    Okay, so a y-intercept is a point where the x value is zero, thus intersecting the y-axis. So this means that our tangent lines have the point (0, -4) on them.

    First, lets find the equation for our tangent lines. We know (or should) that y\prime is our slope, this means that in the equation y=mx+b, m = y\prime

    So first lets find y\prime

    I'm assuming you understand how to differentiate,
    Y = X^{2} +1

    Becomes
    Y\prime = 2X
    If that step confused you, say so. You might also see this written as
    \frac{dy}{dx} = 2x
    They are the same thing.


    So our tangent line has an equation of y=mx+b, and we now know that our slope = 2x for some x. We need to find that x in order to know our slope. We also know that to calculate a slope, we can take the difference in y of two points and divide by the difference in x. Well, we have one point of (0, -4), and we can calculate our other point as ( x, x^{2}+1) because we know that there is an x, y point on the parabola that is the same as on the line.

    So
    \frac{y_1 - y_2}{x_1 - x_2} = slope

    replace the left side with our points, and the right side with 2x, because we know that 2x is our slope.
    \frac{(x^{2}+1) - (-4)}{x - 0} = 2x

    simplify
    \frac{x^{2} + 5}{x} = 2x

    multiply by x
    x^{2} + 5 = 2x^{2}

    subtract x^{2}
    5 = x^{2}

    square root
    \frac{+}{}\sqrt{5} = x

    plug that in for our slope of 2x to get slope = \frac{+}{}2\sqrt{5}

    And now, FINALLY (lol) we can plug that into y=mx+b
    so y = \frac{+}{}2\sqrt{5}x+b

    And plug in our x, y values
    -4 = \frac{+}{}2\sqrt{5}(0)+b
    -4 = b

    And substitute b in to our equation
    y = \frac{+}{}2\sqrt{5}x-4
    Last edited by angel.white; October 18th 2007 at 11:17 PM.
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  6. #6
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    Quote Originally Posted by ruscutie100 View Post
    Find the equations of the tangent lines to Y = X^2 +1 that have a y intercept of -4
    Hello,

    the y-intercept is a point on the y-axis that means the x-value is zero.

    1. Let this point be P(0,-4)
    2. Calculate the first derivative which is the slope of the tangent:

    y=x^2+1~\implies~y'=2x

    3. The tangent point T(t, t^2+1) is placed on the graph of the function too.

    4. Plug in the coordinates of T:

    y'=2t = m

    Use the point-slope-formula of the equation of a straight line:

    2t=\frac{t^2+1-(-4)}{t-0} Solve for t. I've got: t=-\sqrt{5}~\vee~t=\sqrt{5}

    5. Plug in these values:

    a) The tangent points are T_1(-\sqrt{5}, 6) or T_2(\sqrt{5}, 6)

    b) the 2 tangents have the equations:
    \boxed{t_1:y=-2\sqrt{5} \cdot x-4} or \boxed{t_2:y=2\sqrt{5} \cdot x-4}
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  7. #7
    Super Member angel.white's Avatar
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    Okay, sorry that took so long, I had to get some things straight in my head.

    It should be correct now, I won't edit it again.
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