# Thread: Tangent Graph Calc BC Problem

1. ## Tangent Graph Calc BC Problem

Let f(x)= 6-x². For 0 < w < √6 , let A(w) represent the area of the triangle created by the coordinate axes and the line tangent to the graph of f at the point (w, 6-w²). Find A(1).

Here is the Diagram of what I'm attempting to work with

http://img472.imageshack.us/img472/50/a5nt.jpg

A start to this problemw ould greatly be appreciated.

2. Originally Posted by r2d2
Let f(x)= 6-x². For 0 < w < √6 , let A(w) represent the area of the triangle created by the coordinate axes and the line tangent to the graph of f at the point (w, 6-w²). Find A(1).

Here is the Diagram of what I'm attempting to work with

http://img472.imageshack.us/img472/50/a5nt.jpg

A start to this problemw ould greatly be appreciated.

The equation of a line tangent to f at a point b is given by y = f'(b)(x - b) + f(b). So if f = 6 - x^2, then f(1) = 5, f'(1) = -2 and the line is given by y = -2(x - 1) + 5 = -2x + 7. The y-intercept of this line is 7, and the x-intercept is at 7/2. So the area of the triangle is 1/2 * 7 * 7/2 = 49/4 = 12.25.

3. Thank you for your response.

I understand all except that in the tangent line equation, why would the "x" in y=mx+b be (x-b)? That part seems somewhat confusing.

Thanks

And From there, would you plug 1 into the tangent line equation to find A(1), which would be 5?

4. Originally Posted by r2d2