# Equation of tangent line to the graph

• May 31st 2009, 04:53 PM
hemi
Equation 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:
$g(x)=(x+2)(\frac{x-5}{x+1})$ at $(0,-10)$
• May 31st 2009, 06:09 PM
Amer
derive

$g(x)=(x+2)\left(\frac{x-5}{x+1}\right)$

$g'(x)=\left(\frac{x-5}{x-1}\right)+(x+2)\left(\frac{(x+1)-(x-5)}{(x+1)^2}\right)$

substitute 0 to get the slope

$g'(0)=17.....wright??$

now

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

$tangent...line....y+10=17x$
• May 31st 2009, 07:15 PM
hemi
Quote:

Originally Posted by Amer
derive

$g(x)=(x+2)\left(\frac{x-5}{x+1}\right)$

$g'(x)=\left(\frac{x-5}{x-1}\right)+(x+2)\left(\frac{(x+1)-(x-5)}{(x+1)^2}\right)$

substitute 0 to get the slope

$g'(0)=17.....wright??$

now

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

$tangent...line....y+10=17x$

Um isn't g'(x) = $\frac{x-5}{x+1}+\frac{x+2}{x+1}-\frac{(x+2)(x-5)}{(x+1)^2}$
and thus:

g'(0) = 7 ?

who's calculation is right here?
• May 31st 2009, 08:39 PM
mr fantastic
Quote:

Originally Posted by Amer
derive

$g(x)=(x+2)\left(\frac{x-5}{x+1}\right)$

$g'(x)=\left(\frac{x-5}{x {\color{red}+} 1}\right)+(x+2)\left(\frac{(x+1)-(x-5)}{(x+1)^2}\right)$ Mr F says: Typo corrected (in red).

substitute 0 to get the slope

$g'(0)=17.....wright??$ Mr F says: So g'(0) = 7.

[snip]

Quote:

Originally Posted by hemi
Um isn't g'(x) = $\frac{x-5}{x+1}+\frac{x+2}{x+1}-\frac{(x+2)(x-5)}{(x+1)^2}$
and thus:

g'(0) = 7 ?

who's calculation is right here?

Amer made a small typo. It should be obvious how to fix it (even more so if you read the above) and therefore how to get the correct answer.
• May 31st 2009, 08:57 PM
hemi
yeh, it will simply become y+10=7x correct?
• May 31st 2009, 09:23 PM
Amer
ohh
sorry (Worried)