Question

What is the equation of the tangent line of `f(x)=sinx/(x^2-3x+2)` at `x=0`?

Answer 1

`y=1/2x`. See two Socratic graphs, one for zooming the tangent, and the other, for contracting the graph, to reveal branches hidden in the first graph.

Explanation:

graph{(y(x^2-3x+2)-sin x)(y-x/2)=0 [-0.864, 0.764, -0.412, 0.402]} `y = f(0)= 0` at x = 0.

So, the point of contact of the tangent is P( 0, 0 ).

Cross multiply , differentiate and set x = f=0.

`f'(0)(2)+(0)(-3)=cos 0 = 1`, giving the slope `f'(0)=1/2`

And so, the equation of the tangent at P( 0, 0 ) is

`y-0=1/2(x-0)`, giving `y = 1/2x`.

graph{(y(x^2-3x+2)-sin x)(y-x/2)=0 [-26.1, 26, -13.02, 13.02]}