I need to find an equation of a tangent line to the graph of a function at the given point.
Given:
$\displaystyle g(x)=(x+2)(\frac{x-5}{x+1})$ at $\displaystyle (0,-10)$
derive
$\displaystyle g(x)=(x+2)\left(\frac{x-5}{x+1}\right)$
$\displaystyle g'(x)=\left(\frac{x-5}{x-1}\right)+(x+2)\left(\frac{(x+1)-(x-5)}{(x+1)^2}\right)$
substitute 0 to get the slope
$\displaystyle g'(0)=17.....wright??$
now
$\displaystyle slope=\frac{y-y_0}{x-x_0}=\frac{y--10}{x-0}=17$
$\displaystyle tangent...line....y+10=17x$