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I'm stuck on this question:
Find the equations of all lines tangent to y=x^2-4 that pass through the point (5,5).
I found the derivative of the function (y'=2x), and plugged in 5 to get a slope of 10. Then I put it into the y=mx+b form to solve for "b". I got y=10x-45. But when I graph the original function and the equation of the tangent line, the tangent line isn't... tangent. Help, please!
The point (5,5) is not lies on curve y=x^2-4.
I understand that. I have no idea how to relate the two...Quote:
Originally Posted by Also sprach Zarathustra
1. Draw the graph y=x^2-4
2. Draw some tangent line to the graph.
3. Put the point (5,5) on the line.
NOW, think how you can get this tangent line equation!
Hint 1:
Call the tangent point A(x,y)=A(x,x^2-4). You have also the point (5,5) on this line... so you can find the slope(by two points formula).
Thank you!!