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Thread: [SOLVED] Tangent Line/Secant Line To A Parabola

  1. #1
    r4nd0mz
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    [SOLVED] Tangent Line/Secant Line To A Parabola

    You can find the slope of the tangent line to a parabola $\displaystyle y=x^2$ at a point $\displaystyle P(a,a^2)$ using the following method.

    (a) Find an expression, in terms of x and a, for the slope of a secant line from $\displaystyle P$ to another point $\displaystyle Q(x,x^2)$ on the parabola.

    (b) Simplify this slope assuming $\displaystyle x \not= a$. What number, in terms of $\displaystyle a$, does this quantity approach as $\displaystyle x$ approaches $\displaystyle a$?
    Well for part (a) since the equation of a slope is $\displaystyle \frac{y_2-y_1}{x_2-x_1}$ then the expression should be $\displaystyle \frac{a^2-x^2}{a-x}$ right?

    I don't understand part (b) at all.
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  2. #2
    Moo
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    Hello,

    They're asking you to what value the quotient tends when x will be about to equal a. It's called the limit of the expression when x tends to a.

    For this thing, you can factorize (a-x) with the (a-b)(a+b)=a-b formula. And as $\displaystyle x \neq a $, $\displaystyle a-x \neq 0$, you can simplify ;-)

    If you don't understand the notion of limit, it's like replacing x by a.
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    They're asking you to what value the quotient tends when x will be about to equal a. It's called the limit of the expression when x tends to a.

    For this thing, you can factorize (a-x) with the (a-b)(a+b)=a-b formula. And as $\displaystyle x \neq a $, $\displaystyle a-x \neq 0$, you can simplify ;-)

    If you don't understand the notion of limit, it's like replacing x by a.
    Ok but for the second part it said to simplify the slope instead of $\displaystyle x \neq a $. How would I go by simplifying the slope?

    Edit - nevermind. I got the slope to be simplified to $\displaystyle a+x$. But how would go you go about to find the "the limit of expression" when $\displaystyle x$ approaches $\displaystyle a$.
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  4. #4
    Moo
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    When x approaches a, you can say that x+a approaches a+a, which is 2a. And this is what you're asked to get :-)
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  5. #5
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    Quote Originally Posted by Moo View Post
    When x approaches a, you can say that x+a approaches a+a, which is 2a. And this is what you're asked to get :-)
    Ok...I understand now. Thx alot.
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