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 $\displaystyle (0,-10)$

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- May 31st 2009, 04:53 PMhemiEquation of tangent line to the graph
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 $\displaystyle (0,-10)$ - May 31st 2009, 06:09 PMAmer
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

$\displaystyle slope=\frac{y-y_0}{x-x_0}=\frac{y--10}{x-0}=17$

$\displaystyle tangent...line....y+10=17x$ - May 31st 2009, 07:15 PMhemi
- May 31st 2009, 08:39 PMmr fantastic
- May 31st 2009, 08:57 PMhemi
yeh, it will simply become y+10=7x correct?

- May 31st 2009, 09:23 PMAmer
ohh

sorry (Worried)