Thread: Equation of tangent line to the graph

1. 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:
$\displaystyle g(x)=(x+2)(\frac{x-5}{x+1})$ at $\displaystyle (0,-10)$

2. 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

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

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

3. Originally Posted by Amer
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

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

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

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

g'(0) = 7 ?

who's calculation is right here?

4. Originally Posted by Amer
derive

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

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

substitute 0 to get the slope

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

[snip]

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

5. yeh, it will simply become y+10=7x correct?

6. ohh
sorry