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Math Help - Finding the tangent and secants of a point of a given function?

  1. #1
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    Finding the tangent and secants of a point of a given function?

    I am really struggling in understanding this concept (mainly because I'm doing my math course online, and it's pre-calculus and not calculus), and I couldn't find any thing on this concept at all online after multiple tries on different search engines. I know it has something to do with instantaneous velocity (instantaneous rate of change), but my book doesn't explain any further than that.

    If anyone can point me in the direction(helpful links) or explain the concepts to me, I'd greatly appreciate it. Thank you in advance.
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    Post a particular question you are having problems with.
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    Alright, here's the question:
    Answer the following questions, given the equation -3(1/2)^x^-^2+6

    a) Graph the function using a suitable scale and any means at your disposal (this I can do, as I have graphing software).

    b) Draw the secants from y-intercept to each of the points on the curve ending at x=2,1,0.5,0.1,0.01,0.001 (this confuses me, because there were no x and y given for the equation which I would have made equal to x2 and y2).

    c) Determine the equation of each of the secants constructed in part b (I usually do this part before graphing the secants in order to make it easier to do the points).

    d) determine the equation of the tangent at y-intercept in simplified form (I have absolutely no idea on how to answer this one).

    I would greatly appreciate it if you could explain all the steps you took, I kind of understand the concepts so far but not in clarity (I am thinking x = 0, because the secant points are approaching the zero, but I could be wrong).

    Thank you in advance.
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    Quote Originally Posted by Pupil View Post
    b) Draw the secants from y-intercept to each of the points on the curve ending at x=2,1,0.5,0.1,0.01,0.001 (this confuses me, because there were no x and y given for the equation which I would have made equal to x2 and y2).

    c) Determine the equation of each of the secants constructed in part b (I usually do this part before graphing the secants in order to make it easier to do the points).
    Lets start with these two parts. With these points x=2,1,0.5,0.1,0.01,0.001 one at a time compare them to the y-intercept (x=0) like this

    x=0: gives -3(1/2)^{(0)-2}+6=\dots what do you get?

    x=2: gives -3(1/2)^{(2)-2}+6=\dots what do you get?

    x=1: gives -3(1/2)^{(1)-2}+6=\dots what do you get?

    x=0.5: gives -3(1/2)^{(0.5)-2}+6=\dots what do you get?

    and so on....

    then draw the lines connecting

    (0, f(0)) and (2, f(2))

    (0, f(0)) and (1, f(1))

    (0, f(0)) and (0.5, f(0.5))

    and so on...

    Then find the equation of the straight line between each of these pairs.

    What do you notice about the gradient of these lines as x gets very small?
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  5. #5
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    Quote Originally Posted by Pupil View Post
    d) determine the equation of the tangent at y-intercept in simplified form (I have absolutely no idea on how to answer this one).
    Ever heard of differentiation?

    This might help you: Khan academy - Equation of a tangent line
    I find the Khan academy videos for calculus are very educational.

    The y-intercept is when x = 0. To find the equation of the tangent line you would need to find the gradient by differentiation and then find the C constant by using the initial equation.
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    Quote Originally Posted by pickslides View Post
    Lets start with these two parts. With these points x=2,1,0.5,0.1,0.01,0.001 one at a time compare them to the y-intercept (x=0) like this

    x=0: gives -3(1/2)^{(0)-2}+6=\dots what do you get?

    x=2: gives -3(1/2)^{(2)-2}+6=\dots what do you get?

    x=1: gives -3(1/2)^{(1)-2}+6=\dots what do you get?

    x=0.5: gives -3(1/2)^{(0.5)-2}+6=\dots what do you get?

    and so on....

    then draw the lines connecting

    (0, f(0)) and (2, f(2))

    (0, f(0)) and (1, f(1))

    (0, f(0)) and (0.5, f(0.5))

    and so on...

    Then find the equation of the straight line between each of these pairs.

    What do you notice about the gradient of these lines as x gets very small?
    At x=0I get f(0)=-6 Which I've come to find is the original function.

    At x=2 I get f(2)=3 So this would be my first secant, correct? x2=0, x1=2, y2=-6, y1=3? And then rinse and repeat for the rest of the secants, connecting them with the original point (0,-6)? So the equation for the first secant would be 4.5x-6 right?


    I notice that, while x may be gradually decreasing in value as it approaches (0,-6) but the slope increases in value.
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    Quote Originally Posted by Educated View Post
    Ever heard of differentiation?

    This might help you: Khan academy - Equation of a tangent line
    I find the Khan academy videos for calculus are very educational.

    The y-intercept is when x = 0. To find the equation of the tangent line you would need to find the gradient by differentiation and then find the C constant by using the initial equation.
    I watched the video and it seems too calculus heavy on me for now, he applied concepts and terms that are foreign to me. Do I need a solid understanding of calculus in order to get differentiation, or could I learn it now without knowing anything about calculus yet?
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    Quote Originally Posted by Educated View Post
    Ever heard of differentiation?

    This might help you: Khan academy - Equation of a tangent line
    I find the Khan academy videos for calculus are very educational.
    It seems to me that this problem was intended as an introduction to the concept of differentiation and that, therefore, using differentiation to find the tangent line is not intended. pickslide's method is best- actually find the equation of each of those tangent lines and look what happens to the slope as the "second" point moves closer to the first point.
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  9. #9
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    Quote Originally Posted by HallsofIvy View Post
    It seems to me that this problem was intended as an introduction to the concept of differentiation and that, therefore, using differentiation to find the tangent line is not intended. pickslide's method is best- actually find the equation of each of those tangent lines and look what happens to the slope as the "second" point moves closer to the first point.
    Ah, so the equation would then be an approximation? So to find the equation of the tangent, I would have to find the equation of the closest secant to that point of the function?

    Thanks for all the help guys, I appreciate it.
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