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Thread: Help with finding secant and tangent gradient?

  1. #1
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    Help with finding secant and tangent gradient?

    I'm new to calculus and having a lot of trouble with this question.

    Curve y=2x^2-5 has two points P and Q. The abscissa of P is x and Q is x+h.
    a) Find the coordinates of P and Q.
    b) Using these find the gradient of the secant PQ.
    c) Find the gradient of the tangent at P by letting h approach 0.

    I've tried different methods but can't seem to pull the correct answer out. Any help would be greatly appreciated
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  2. #2
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    Re: Help with finding secant and tangent gradient?

    part (a) ...

    coordinates of point $P$ are $(x,2x^2-5)$

    coordinates of point $Q$ are $(x+h,2(x+h)^2-5)$

    part (b) ...

    $\dfrac{\Delta y}{\Delta x}=\dfrac{[2(x+h)^2-5]-(2x^2-5)}{(x+h)-x}$

    part (c) ...

    simplify the difference quotient from part (b), then determine the limit as $h \to 0$

    You've stated you have tried "different methods", try this and post up your results so someone may provide further guidance.
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  3. #3
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    Re: Help with finding secant and tangent gradient?

    Then show us what you have tried and perhaps we can point out any errors. (a) and (b) at least follow from the basic definitions. The coordinates of P are (x, 2x^2- 5). The coordinates of Q are (x+h, 2(x+ h)^2- 5). The "gradient of the secant PQ" is, by definition of "gradient" (2(x+ h)^2- 5- (2x^2- 5))/(x+h- x). What do you get for 2(x+ h^2- 5? What did you get for 2(x+ h)^2- 5+ 2x^2+ 5?

    Do you see that you can cancel "h" in numerator and denominator of that "gradient" and can then simply set h equal to 0?
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