Question

tangents!! tangents!! @Mathematics

Answer 1

\[f(x)=\frac{-8x}{(x^2+1)}\]The derivative of this function at the point (0,0) will be the slope of the tangent line,m, at this point. We have:\[f'(x)=\frac{-8(x^2+1)+16x^2}{(x^2+1)^2}=\frac{8x^2-8}{(x^2+1)^2}\]At x=0 we get:\[f'(0)=-8\]So an equation of the tangent line at (0,0) in point-slope form is:
\[y-0=-8(x-0)\]Or,\[y=-8x\]

Answer 2

a is correct.

Answer 3

At (1,-4) we have:\[f'(1)=0\]So we have a horizontal tangent here, the equation of which is:\[y-1=0(x+1)\]Or,\[y=1\]

Answer 4

b is incorrect.

Answer 5

Not seeing my error...

Answer 6

then?

Answer 7

lol
I see it:
y+4=0(x-1)
y=-4

Answer 8

ooops lol