absolute value
at x=2, the absolute value thing is positive
do this one
[tex]y=x \sqrt{5-x^2} [/tex]
the derivitive is
[tex] \frac{dy}{dx}=\sqrt{5-x^2}- \frac{x^2}{ \sqrt{5-x^2} } [/tex]
so the slope at x=2 is
[tex] \frac{dy}{dx}=\sqrt{5-2^2}- \frac{2^2}{ \sqrt{5-2^2} } [/tex]=-3
a point is (2,2)
point slope form
y-y1=m(x-x1)
y-2=-3(x-2)
or
y=-3x+8
das is de equation
