Tangent line help(without calculus)

Question

I need to find to find a tangent line to the curve $x \over {x^2 + x + 2}$.

Answer 1

I suppose the "without Calculus" was added after the comment and answer you have already received. This is "almost" Calculus: $f(x)= \frac{x}{x^2+ x+ 2}$ and, for any h,$f(x+ h)= \frac{x+ h}{(x+ h)^2+ (x+ h)+ 2}= \frac{x+ h}{x^2+ 2xh+ h^2+ x+ h+ 2}$.

The difference is $\frac{x+ h}{x^2+ (2h+1)x+ h^2+ h+ 2}- \frac{x}{x^2+ x+ 2}= \frac{(x+ h)(x^2+ x+ 2)- x(x^2+ (2h+1)x+ h^2+ h+ 2}{(x^2+ x+ 2)(x^2+ (2h+1)x+ h^2+ h+ 2)}$$= \frac{x^3+ x^2+ 2x+ hx^2+ hx+ 2h- x^3- (2h+1)x^2- h^2x- hx- 2h}{(x^2+ x+ 2)(x^2+ (2h+1)x+ h^2+ h+ 2)}$$= \frac{hx^2+ (h^2-h)x- 2h}{(x^2+ x+ 2)(x^2+ (2h+1)x+ h^2+ h+ 2)}= \frac{h(x^2+ (h- 1)x- 2)}{(x^2+ x+ 2)(x^2+ (2h+1)x+ h^2+ h+ 2)}$. Dividing by h gives the "difference quotient" $\frac{(x^2+ (h- 1)x- 2)}{(x^2+ x+ 2)(x^2+ (2h+1)x+ h^2+ h+ 2)}$. That is the slope of the line through the two points (x,f(x)) and (x+h, f(x+h)). Taking the limit as h goes to 0 gives the slope of the tangent line, $\frac{x^2- x- 2}{(x^2+ x+ 2)^2}$.

The equation of the tangent line through $(x_0, f(x_0))$ is $y= \frac{x_0^2- x_0- 2}{(x_0^2+ x_0+ 2)^2}(x- x_0)+ f(x_0)$

Answer 2

Let's find the tangent to the graph $y=f(x)$ of $f(x)=\frac{x}{x^2+x+2}$ at $x=t$;

To use the tangent cone method we translate $(t,f(t))$ to the origin:

$y=-t/(t^2+t+2)+\frac{x+t}{(x+t)^2+(x+t)+2)}$

which simplifies to

$y((t^2+t+2)x^2+(2t^3+3t^2+5t+2)x+t^4+2t^3+5t^2+4t+4)+(tx^2+(t^2-2)x)=0$

from which we take the tangent cone (skip all terms of higher degree in $x$ and $y$):

$y (t^2+t+2)^2+(t^2-2)x=0$

or

$y = -\frac{t^2-2}{(t^2+t+2)^2} x$.

Now $-\frac{t^2-2}{(t^2+t+2)^2}$ is the slope of the shifted curve at the origin which corresponds to the slope of the original curve at $x=t$.

So the tangent is

$y=-\frac{t^2-2}{(t^2+t+2)^2}(x-t)+\frac{t}{t^2+t+2}$