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

You can find the slope of the tangent line to a parabola at a point using the following method.

(a) Find an expression, in terms of x and a, for the slope of a secant line from to another point on the parabola.

(b) Simplify this slope assuming . What number, in terms of , does this quantity approach as approaches ?

Well for part (a) since the equation of a slope is then the expression should be right?

I don't understand part (b) at all.

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 (a²-x²) with the (a-b)(a+b)=a²-b² formula. And as , , you can simplify ;-)

If you don't understand the notion of limit, it's like replacing x by a.

Originally Posted by Moo

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 (a²-x²) with the (a-b)(a+b)=a²-b² formula. And as , , 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 . How would I go by simplifying the slope?

Edit - nevermind. I got the slope to be simplified to . But how would go you go about to find the "the limit of expression" when approaches .

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 :-)

Originally Posted by Moo

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.