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Thread: Equation of Secant Line

  1. #1
    Super Member harpazo's Avatar
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    Equation of Secant Line

    Given f(x) = -3x + 2, find the equation of the secant line at x = 1 with h = 0.01.

    The book instructs students to use

    M_sec = [f(x + h) - f(x)]/h

    I have seen this expression before. This is the difference quotient of f, is it not?

    I am seeking the needed steps that will guide me through this problem.
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    Re: Equation of Secant Line

    Did you do any "googling"?
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    Re: Equation of Secant Line

    Quote Originally Posted by harpazo View Post
    Given f(x) = -3x + 2, find the equation of the secant line at x = 1 with h = 0.01.

    The book instructs students to use

    M_sec = [f(x + h) - f(x)]/h

    I have seen this expression before. This is the difference quotient of f, is it not? I haven't heard it called that, but it is simply the gradient of the line required.

    I am seeking the needed steps that will guide me through this problem.
    The secant line joins two points on the curve. One point is (1, f(1)) and the other is (1.01, f(1.01)).

    You now need to find the equation of the line through these two points n the usual way.
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    MHF Contributor MarkFL's Avatar
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    Re: Equation of Secant Line

    Just going by intuition first, what do you suspect the secant line to a linear function is going to be?
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    Re: Equation of Secant Line

    Quote Originally Posted by MarkFL View Post
    Just going by intuition first, what do you suspect the secant line to a linear function is going to be?
    Oh, of course, didn't take any notice of f(x).
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    Re: Equation of Secant Line

    Quote Originally Posted by DenisB View Post
    Did you do any "googling"?
    No.
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    Re: Equation of Secant Line

    Quote Originally Posted by MarkFL View Post
    Just going by intuition first, what do you suspect the secant line to a linear function is going to be?
    Perpendicular.
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    Re: Equation of Secant Line

    Quote Originally Posted by harpazo View Post
    Perpendicular.
    No.

    Do you know what the secant of a function looks like? (Hint: Think of the secant of a circle.)

    I don't mean to criticize but I really don't think you are ready for this kind of problem. This is Pre-Calculus and you've had a lot of problems with Algebra. I'd advise more Algebra review.

    -Dan
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    Super Member harpazo's Avatar
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    Re: Equation of Secant Line

    Quote Originally Posted by topsquark View Post
    No.

    Do you know what the secant of a function looks like? (Hint: Think of the secant of a circle.)

    I don't mean to criticize but I really don't think you are ready for this kind of problem. This is Pre-Calculus and you've had a lot of problems with Algebra. I'd advise more Algebra review.

    -Dan
    This problem is from my Michael Sullivan College Algebra 9th Edition textbook. I think I can do it. Thanks anyway.
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    MHF Contributor MarkFL's Avatar
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    Re: Equation of Secant Line

    Let's look at the definition of a secant line:

    In geometry, a secant of a curve is a line that intersects the curve in at least two points. The word secant comes from the Latin word secare, meaning to cut.

    So, imagine a line in the plane, and then further imagine two points on that line and a line though those two points. Can you see that the secant line will coincide with the original line?

    Consider the linear function:

    $\displaystyle f(x)=ax+b$

    Now, consider two arbitrary points on this linear function:

    $\displaystyle (x,f(x))=(x,ax+b)$

    $\displaystyle (x+h,f(x+h))=(x+h,a(x+h)+b)$

    And so the slope $\displaystyle m$ of the secant line is given by:

    $\displaystyle m=\frac{f(x+h)-f(x)}{(x+h)-(x)}=\frac{(ax+ah+b)-(ax+b)}{h}=a$

    Then using the point slope formula, we find the equation of the secant line to be:

    $\displaystyle y=a(x-x)+f(x)=f(x)$

    Thus, we find the secant line does in fact coincide with the original line.
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    Super Member harpazo's Avatar
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    Re: Equation of Secant Line

    Quote Originally Posted by MarkFL View Post
    Let's look at the definition of a secant line:

    In geometry, a secant of a curve is a line that intersects the curve in at least two points. The word secant comes from the Latin word secare, meaning to cut.

    So, imagine a line in the plane, and then further imagine two points on that line and a line though those two points. Can you see that the secant line will coincide with the original line?

    Consider the linear function:

    $\displaystyle f(x)=ax+b$

    Now, consider two arbitrary points on this linear function:

    $\displaystyle (x,f(x))=(x,ax+b)$

    $\displaystyle (x+h,f(x+h))=(x+h,a(x+h)+b)$

    And so the slope $\displaystyle m$ of the secant line is given by:

    $\displaystyle m=\frac{f(x+h)-f(x)}{(x+h)-(x)}=\frac{(ax+ah+b)-(ax+b)}{h}=a$

    Then using the point slope formula, we find the equation of the secant line to be:

    $\displaystyle y=a(x-x)+f(x)=f(x)$

    Thus, we find the secant line does in fact coincide with the original line.
    Nicely-done!
    Thanks from MarkFL
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    Re: Equation of Secant Line

    Quote Originally Posted by harpazo View Post
    This problem is from my Michael Sullivan College Algebra 9th Edition textbook. I think I can do it. Thanks anyway.
    No offense intended, but I do believe a College Algebra text is beyond you in many topics. In many respects College Algebra and Pre-Calculus overlap. Still, let's give it a try.

    -Dan
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    Re: Equation of Secant Line

    Quote Originally Posted by Debsta View Post
    The secant line joins two points on the curve. One point is (1, f(1)) and the other is (1.01, f(1.01)).

    You now need to find the equation of the line through these two points n the usual way.
    Just repeating my earlier comment above.

    If the function is a straight line, then the secant (the straight line joining any two points on the curve think about it!) must be the same straight line.

    Doing it the long way and using the method implied in the question "Given f(x) = -3x + 2, find the equation of the secant line at x = 1 with h = 0.01":

    If x=1, y= -3*1+2 = -1, so one point is (1, -1).

    If x= 1.01, y = -3*1.01 + 2 = -1.03 , so other point is (1.01, -1.03)


    Gradient of secant = (-1.03 +1)/(1.01 - 1) =-0.03/0.01 = -3


    Using y - y1 = m(x - x1) where m = -3 and (x1, y1)=(1, -1):


    y + 1 =-3 (x-1)

    y = -3x +3 -1

    y=-3x +2 (confirming that the secant is the same line (obviously))


    This is the method to use if the original function is not a straight line.
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  14. #14
    Super Member harpazo's Avatar
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    Re: Equation of Secant Line

    Quote Originally Posted by Debsta View Post
    Just repeating my earlier comment above.

    If the function is a straight line, then the secant (the straight line joining any two points on the curve think about it!) must be the same straight line.

    Doing it the long way and using the method implied in the question "Given f(x) = -3x + 2, find the equation of the secant line at x = 1 with h = 0.01":

    If x=1, y= -3*1+2 = -1, so one point is (1, -1).

    If x= 1.01, y = -3*1.01 + 2 = -1.03 , so other point is (1.01, -1.03)


    Gradient of secant = (-1.03 +1)/(1.01 - 1) =-0.03/0.01 = -3


    Using y - y1 = m(x - x1) where m = -3 and (x1, y1)=(1, -1):


    y + 1 =-3 (x-1)

    y = -3x +3 -1

    y=-3x +2 (confirming that the secant is the same line (obviously))


    This is the method to use if the original function is not a straight line.
    Thank you for your kindness and patience.
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