find the equation of the tangent and normal to each of the following curves, the lengths of the subtangent and subnormal, then trace the curve showing these lines.

y= x^2 - 6x + 4 at (4, -4) ANSWERS ARE: 2X-Y-12=0 AND X+2Y +4 =0.

Question

find the equation of the tangent and normal to each of the following curves, the lengths of the subtangent and subnormal, then trace the curve showing these lines.

y= x^2 - 6x + 4 at (4, -4) ANSWERS ARE: 2X-Y-12=0 AND X+2Y +4 =0.

anonymous · Answer

help, my teacher taught us a strategy to answer this by using calculus.. a basic one

anonymous · Answer

You know that a linear function such as a tangent or a normal can be written on the form of $$y(x)=kx+m$$
If you take the derivative of your function y(x) (the one you wrote out); you'll get the function y'(x). The thing about this function is that it gives you the k-value in the above function in a certain point (you simply insert a certain x-value into y'(x) ).

anonymous · Answer

so the derivative of y= x^2 - 6x + 4 is 2x-6?

anonymous · Answer

Indeed.

anonymous · Answer

Basically what I'm saying -- if you didn't get it the first time -- is that the slope of your function y in a point x is the same thing as the derivative IN that point.

anonymous · Answer

Meaning the k-value for your tangent in a point b is the same thing as y'(b)=''blabla''
(Man am I repeating myself, hehe!).

anonymous · Answer

got it now!! thanks!! @Noliec